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UNIVERSITY  OF  CALIFORNIA 

ANDREW 

SMITH 

HALLIDIE: 

1901 


ELECTRICAL  PROBLEMS 


FOR 


ENGINEERING  STUDENTS 


BY 

WILLIAM   L.  HOOPER,  PH.  D. 

PROFESSOR  OF  ELECTRICAL  ENGIKEERING,  TUFTS  COLLEGE,  MASS. 


ROY  T.  WELLS,  M.S. 

SENIOR  FELLOW  ix  PHYSICS,  CLARK  UNIVERSITY,  WORCESTER,  MASS. 


BOSTON,  U.S.A.,  AND  LONDON 
GINN   &   COMPANY,  PUBLISHERS 


1902 


HALLIDIE 


ENTERED  AT  STATIONERS'  HALL 


COPYRIGHT,  1902,  BY 
WILLIAM  L.  HOOPER  AND  ROY  T.  WELLS 


ALL  RIGHTS  RESERYED 


PREFACE 

"  Electrical  Problems "  has  been  written  in  the  belief  that 
there  is  a  field  and  should  be  a  demand  for  this  class  of  text- 
book. 

In  physical  science  generally,  and  especially  in  engineering, 
knowledge  is  of  use  only  so  far  as  we  are  able  to  calculate 
numerical  results,  and  ability  to  obtain  such  results  quickly  and 
accurately  is  to  be  obtained  only  by  extensive  practice.  Then, 
too,  the  ordinary  mind  arrives  at  a  clear  conception  of  general 
principles  only  by  the  way  of  concrete  examples.  To  the  aver- 
age student  mathematical  formulae  are  vague  and  uninviting 
until  he  has  himself  made  practical  application  of  them. 

For  some  years  the  students  of  electrical  engineering  in  Tufts 
College  have  been  largely  exercised  in  the  solution  of  numerical 
problems,  and  both  the  experience  of  the  instructors  and  the 
testimony  of  the  students  themselves  clearly  indicate  the  value 
of  such  work  in  rendering  clear  and  precise  their  views  of  elec- 
trical phenomena.  Especially  valuable  in  clarifying  the  students' 
conceptions  of  physical  relations  are  those  problems  whose 
answers  appear  in  the  form  of  curves,  showing  the  effects  of 
varying  the  quantities  involved. 

Most  of  the  problems  in  this  book  have  already  been  presented 
to  the  electrical  classes  in  Tufts  College.  Nevertheless,  among 
so  large  a  number  it  is  probable  that  errors,  especially  in  the 
answers,  remain  to  be  corrected.  Those  communicating  such 
corrections  to  them  will  receive  the  hearty  thanks  of 

THE    AUTHORS. 

TUFTS  COLLEGE, 

September,  1902. 

iii 


CONTENTS 


CHAPTER  PAGE 

I.    THE  FUNDAMENTAL  UNITS 1 

II.    CURRENT,   ELECTRO-MOTIVE    FORCE,   AND   RESISTANCE  4 

III.  RESISTIVITY 7 

IV.  ELECTRO-MOTIVE  FORCES  AND  RESISTANCES  IN  SERIES 

GROUPING 11 

V.    ELECTRO-MOTIVE  FORCES  AND  RESISTANCES  IN  PARAL- 
LEL GROUPING 16 

VI.   DROP 21 

VII.   POWER  AND  EFFICIENCY 25 

VIII.    THE  MAGNETIC  FIELD  DUE  TO  A  CURRENT     ....  30 

IX.  INDUCTANCE " 38 

X.    THE  CONDENSER 45 

XI.    THERMOELECTRICITY 51 

XII.    ELECTRO-CHEMISTRY  . 55 

XIII.  ALTERNATING     ELECTRO-MOTIVE     FORCES    AND    CUR- 

RENTS       58 

XIV.  COMBINATION      OF      ALTERNATING      ELECTRO-MOTIVE 

FORCES *  .  64 

XV.    COMBINATION  OF  ALTERNATING  CURRENTS      ....  69 

XVI.   IMPEDANCE 73 

XVII.    DIRECT  CURRENT  ARMATURES 85 

XVIII.    ALTERNATING  CURRENT  ARMATURES 90 

XIX.    THE  WINDING  OF  ARMATURES 94 

XX.   ARMATURE  REACTIONS 107 

XXI.    FIELD  WINDING 112 

XXII.    THE  TRANSFORMER 118 

XXIII.  THE  ROTARY  CONVERTER 124 

XXIV.  THE  INDUCTION  MOTOR 127 

XXV.   TESTING  OF  DYNAMOS 132 

XXVI.    THE  TRANSMISSION  OF  POWER 138 

APPENDIX 141 

WIRE  TABLE 142 

ANSWERS 145 

GRAPHICAL  SOLUTIONS 157 


ELECTRICAL  PROBLEMS 

CHAPTER   I 
THE   FUNDAMENTAL   UNITS 

The  system  of  units  generally  used  in  physical  science 
employs  the  centimeter,  the  gram,  and  the  second,  respectively, 
as  the  units  of  length,  mass,  and  time,  and  hence  is  known 
as  the  centimeter-gram-second,  or  more  briefly  as  the  C.G.S., 
system. 

Since,  however,  commercial  measurements  are  frequently 
made  in  terms  of  other  units,  as,  for  instance,  feet,  pounds,  and 
minutes,  it  is  important  that  one  should  be  able  to  readily 
change  from  one  system  of  units  to  the  other. 

It  is  assumed  that  the  student  is  familiar  with  both  the  metric 
and  the  English  systems.  He  should  remember  that 

2.54  centimeters  =  1  inch  ; 

453.6  grams  =  1  pound  avoirdupois. 

PROBLEMS 

1.  How  many  centimeters  in  a  foot? 

2.  How  many  meters  in  a  mile  ? 

3.  How  many  centimeters  in  a  knot? 

4.  A  No.  6  wire  (Brown  and  Sharp  Gauge,  or  B.S.G.)  has 
a  diameter  of  162  mils.     What  is  its  diameter  in  centimeters? 

XOTE.  —  The  mil  is  the  thousandth  part  of  an  inch. 

1 


2  ELECTRICAL  PROBLEMS 

5.    A  No.  10  wire,  B.S.G.,  is  .102  inch  in  diameter.      What 
is  its  cross  section  in  square  mils  and  in  circular  mils  ? 


E.  —  A  circular  mil  is  the  area  of  a  circle  of  1  mil  diameter;  a  square 
mil  is  the  area  of  a  square  of  1  rail  side.  The  area  in  circular  mils  is 
thus  the  square  of  the  diameter  in  mils,  whereas  the  area  in  square  mils 
is  the  square  of  the  diameter  in  mils  multiplied  by  2£,  or  .7854:.  Thus  the 
cross  section  of  a  wire  of  circular  section  .01  inch  in  diameter  is  100 
circular  mils,  or  78.54  square  mils. 

6.  How  many  circular  mils  in  the  cross  section  of  a  wire 
1.1683  centimeters  in  diameter? 

7.  What  is  the  diameter  in  millimeters  of  a  wire  having 
a  cross  section  of  1256.64  square  mils? 

8.  How  many  cubic  centimeters  in  a   foot  of  a  500,000 
circular  mils  wire? 

9.  What  is   the   cross    section  in  square   millimeters  of   a 
1,000,000  circular  mils  wire? 

10.  How  many  cubic  centimeters  in  a  mile  of  wire  .1  inch  in 
diameter  ? 

11.  How  many  grains  in  a  gram? 

12.  How  many  grains  in  an  ounce  avoirdupois  ? 

13.  One  kilogram  per  kilometer  is  how  many  pounds  per 
1000  feet? 

14.  One  pound  per  1000  feet  is  how  many  kilograms  per 
kilometer  ? 

15.  A  cubic  centimeter  of  water  weighs  1  gram.     Find  the 
weight  in  pounds  of  a  cubic  foot. 

16.  The  specific  gravity  of  drawn  copper  is  8.9.     What  is 
the  weight  of  1000  feet  of  a  wire  102  mils  in  diameter? 

17.  How  long  a  wire  1  millimeter  in  diameter  can  be  drawn 
from  1  pound  of  copper? 

18.  How  long  a  wire  1  mil  in  diameter  can  be  drawn  from 
1  ounce  of  copper? 

19.  What  is  the  weight  in  pounds  per  mile  of  a  copper  wire 
0253  inch  in  diameter  ? 


THE   FUNDAMENTAL   UNITS  3 

20.  If  1000  feet  of  copper  wire  460  mils  in  diameter  weigh 
641  pounds,  what  is  the  weight  of  1  mile  of  copper  wire  128 
mils  in  diameter  ? 

21.  If  a  kilometer  of  copper  wire  .6438  millimeter  in  diam- 
eter weighs  2.89  kilograms,  what  is  the  weight  in  pounds  of 
1  mile  of  copper  wire  11.3  mils  in  diameter? 


CHAPTER    II 
CURRENT,  ELECTRO-MOTIVE  FORCE,  AND  RESISTANCE 

The  three  electrical  units  most  frequently  employed  are  those 
of  current,  electro-motive  force,  and  resistance. 

By  Ohm's  law  the  resistance  of  a  conductor  through  which 
a  uniform  current  is  flowing  is  expressed  by  the  ratio  of  the 
electro-motive  force  to  the  current,  whatever  the  system  of 
units  employed.  In  the  practical  system  the  unit  of  resistance 
is  called  the  ohm,  the  unit  of  electro-motive  force  the  volt,  nnd 
the  unit  of  current  the  ampere. 

For  purposes  of  computation  Ohm's  law  may  be  expressed 
by  any  one  of  the  following  equations  : 


PROBLEMS 

1.  What  current  is  produced  through  a  resistance  of  5  ohms 
by  an  electro-motive  force  of  10  volts? 

2.  What  current  is  produced  through  a  resistance  of  48  ohms 
by  an  electro-motive  force  of  12  volts? 

3.  Through  what  resistance  will  an  electro-motive  force  of 
15  volts  produce  a  current  of  3  amperes  ? 

4.  Through    what    resistance    will    an    electro-motive    force 
of  2.5  volts  produce  a  current  of  10  amperes? 

5.  What    electro-motive    force    will    produce    a    current    of 
4  amperes  through  a  resistance  of  4  ohms  ? 

6.  What   electro-motive    force    will    produce    a    current    of 
.03  ampere  through  a  resistance  of  1000   ohms? 

4 


CURRENT,  ELECTRO-MOTIVE  FORCE,  RESISTANCE     5 

7.  Through    what   resistance  will    121   volts    produce    11 
amperes  ? 

8.  What    electro-motive    force    will    produce    .65    ampere 
through    13    ohms    resistance  ? 

9.  What  current  is  produced  by  10  volts  acting  through 
.25  ohm? 

10.  A  dynamo  generates  500  volts;   the  resistance  of  the  cir- 
cuit, including  the  dynamo,  is  20  ohms.     What  is  the  current  ? 

11.  A  16  candle  power  lamp  requires  .5  ampere,  and  its  resist- 
ance at  full  candle  power  is  200  ohms.    What  voltage  must  be 
impressed  on  its  terminals  ? 

12.  What  electro-motive  force  must  a  dynamo  generate  to 
supply  an  electroplating  current  of  20  amperes  through  a  cir- 
cuit whose  total  resistance  is  .2  ohm  ? 

13.  The  electro-motive  force  supplying  a  current  of  2  amperes 
is  40  volts.     Determine  the  resistance  of  the  circuit. 

14.  A  djmamo  generates  an  electro-motive  force  of  1150  volts 
and  delivers  a  current  of  20   amperes.     The  resistance  of  the 
circuit,  including  dynamo,  is  what? 

15.  A  battery  cell  gives  an  electro-motive  force  of  .9  volt, 
and  has  an  internal  resistance  of  1.2  ohms.     If  its  terminals  are 
short  circuited   by  a  negligible  resistance,  what  current  will 
flow  ? 

16.  An  electric  bell  has  a  resistance  of  400  ohms  and  will 
not  ring  with  a  current  of  less  than  .03  ampere.     Neglecting 
battery  and  line  resistance,  what  is  the  smallest  electro-motive 
force  that  will  ring  the  bell? 

17.  At  20°  C.  No.  10  copper  wire,  B.S.G.,  has  a  resistance 
of  .994  ohm  per  1000  feet.     An  electro-motive  force  of  200 
volts  is  impressed  on  a  circuit  of  it  200  miles  long.     What 
current  will  flow  ? 

18.  What  is  the  resistance  per  mile  of  No.  20  B.S.G.  copper 
wire  at  75°  F.  if  a  current   of   5   amperes   is   passed  through 
100  feet  of  it  and  the  reading  of  a  voltmeter  connected  to  the 
ends  of  the  wire  is  5.13  volts? 


6  ELECTRICAL  PROBLEMS 

19.  If  a  car  heater  is  supplied  with  an  electro-motive  force 
of  500  volts  from  the  trolley,  how  great  must  its  resistance  be 
that  the  current  may  not  exceed  5  amperes  ? 

20.  A  5000  ohm  galvanometer  has  its  terminals  connected 
to  two  points  in  a  circuit  between  which  there  is  a  difference 
of   potential    of    .01   volt.     What   current   flows   through    the 
galvanometer  ?• 

21.  What  electro-motive  force  is  needed  for  an  incandescent 
lamp  of  50  ohms  resistance,  through  which  flows  a  current  of 
1.04  amperes? 

22.  A  battery  whose  electro-motive  force  is  100  volts  is  used 
to  light  a  series  of  4  lamps  having  a  total  resistance  of  33£  ohms. 
What  current  will  flow? 

23.  If  the   magneto-generator  for  ringing  a  telephone  bell 
gives  an  electro-motive  force  of  50  volts,  what  current  will  be 
transmitted  through  the  circuit  if  the  resistance  of  generator 
line  and  bell  is  12,700  ohms? 

24.  What  is  the  maximum  possible  current  from  a  battery 
cell  of    2.1   volts    electro-motive   force  and   .03  ohm    internal 
resistance  ? 

25.  At  0°  C.  No.  9  B.S.G.  copper  wire  has  a  resistance  of 
.739  ohm  per  1000  feet.     A  telephone  trunk  line  of  this  wire 
is  blown  over  on  a  trolley  wire,  short  circuiting  the  trolley  and 
the  rail  with  a  length  of  30  feet.     The  telephone  wire  has  a 
conductivity  98fo  that  of  pure  copper.      Supposing  the  potential 
difference  between  the  trolley  and  rail  to  be  maintained  at  500 
volts,  what  current  will  flow  in  the  No.  9  wire  ? 


CHAPTER    III 
RESISTIVITY 

The  resistivity,  formerly  called  the  specific  resistance,  of  a 
substance  is  the  resistance  between  opposite  ends  of  a  standard 
volume. 

Physicists  usually  express  resistivity  in  ohms  per  cubic 
centimeter  at  the  temperature  of  0°  C.  Thus  the  resistivity 
of  a  sample  of  copper  in  centimeter  ohm -measure  is  the  resist- 
ance in  ohms  at  0°  C.  between  opposite  faces  of  a  cube  of 
this  particular  copper,  each  of  whose  edges  is  1  centimeter 
long.  Engineers  frequently  employ  the  circular  mil-foot  as  a 
standard  volume,  in  which  case  the  resistivity  is  the  resistance 
between  opposite  ends  of  a  wire  of  circular  section,  1  foot 
long  and  .001  inch  in  diameter. 

Other  standard  volumes  are  used,  as  the  meter-millimeter, 
the  meter-gram,  etc. ;  also  resistivities  are  sometimes  expressed 
in  absolute  units  of  .resistance  —  billionths  of  an  ohm  —  per 
cubic  centimeter  when  that  unit  is  employed  as  the  standard 
volume.  The  latter  unit  of  resistance  might  also  be  employed 
with  the  other  standard  volumes. 

The  resistance  of  any  portion  of  a  circuit  uniform  in  material 
and  of  constant  cross  section  equals  the  product  of  its  length 
and  resistivity  divided  by  its  area. 

Generally  an  increase  of  temperature  causes  an  increase  of 
resistivity  in  metals. 

In  most  pure  metals  the  amount  of  this  increase  is  about  1  Jo 
for  every  2.5°  C.  increase  at  and  near  0°  C. 

Hence  the  temperature  coefficients  of  pure  metals  are  approxi- 
mately .004. 

7 


8  ELECTRICAL  PROBLEMS 

PROBLEMS 

1.  The  resistance  of  500  meters  of  copper  wire  with  a  cross 
section  of  .001  square  centimeter  is  79.5  ohms  at  0°  C.  What 
is  the  resistivity  of  copper  in  centimeter  ohm  measure  ? 

Solution.  resistivity  x  length 

resistance  =  — 


area 


resistance  x  area      70.5  x  .001 

resistivity  =  -  =    .n-  -  =  .00000159. 

length  oOO  x  100 

2.  What  is  the  resistivity  in  circular  mil-foot  ohm  measure 
of  copper  ? 

3.  At  60°  C.  what  is  the  resistance  *of  1  mile  of  copper  wire 
42  mils  in  diameter? 

4.  At  77°  F.  how  many  feet  of  No.  10  copper  wire,  B.S.G., 
will  have  a  resistance  of  1  ohm  ?     (See  Appendix.) 

5.  If  the  temperature  in  Problem  4  is  reduced  to  32°  F.,  by 
what  per  cent  must  the  length  of  wire  be  increased  to  still  have 
a  resistance  of  1  ohm  ? 

6.  A  column  of  mercury  106.3  centimeters  long,  with  a  sec- 
tion 1  millimeter  square,  has  a  resistance  of  1  ohm  at  0°  C. 
What  is  the  resistance  between  opposite  faces  of  an  inch  cube  ? 

7.  A  mercury  column,  is  4  millimeters  long  and  6  by  8  milli- 
meters in  section.     What  is  its  resistance  at  0°  C.  ? 

8.  A  battery  gives  an  electro-motive  force  of  208  volts  and 
has  a  negligible  resistance.     The  external  circuit  is  a  coil  of 
10,000  turns  of  copper  wire  10  mils  in  diameter,  the  average 
length  of  a  turn  being  8  inches.     What  current  will  flow  with 
the  copper  at  a  temperature  of  0°  C.? 

9.  If  the  copper  wire  of  Problem  8  is  heated  to  100°  C.,  by 
what  per  cent  will  the  current  be  lessened  ? 

10.  At  0°  C.  1037.3  centimeters  of  iron  wire,  with  a  section 
of  1  square  millimeter,  has  a  resistance  of  1  ohm.     What  is  the 
resistivity  of  iron  in  centimeter  ohm  measure  ? 

11.  What  is  the  resistivity  of  iron  in  circular  mil-foot  ohm 
measure  ? 


RESISTIVITY  9 

12.  The  tube  of  a  mercury  thermometer  has  a  cross  section 
of  1   square  millimeter,  the   bulb  being  of  such  size  that  the 
column  ascends  1  centimeter  for  2°  rise  in  temperature.     The 
lowest  reading  is  —  40°  C.     One  terminal  of  a  circuit  is  fixed 
at  this  point;  the  other  moves  so  as  to  be  always  at  the  top  of 
the  column.     If  the  temperature  coefficient  for  the  resistance 
of  mercury  is  .00072  per  degree  C.,  what  is  the  ratio  of  the 
resistances  of  the  column  at  the  temperatures  of  freezing  and 
boiling  water? 

13.  The  resistance  of  a  1  inch  cube  of  lead  between  opposite 
faces  is   .00000773  ohm.     A  roof  of  lead  is   1000  feet  long, 
20  feet  wide,  and  T^  inch  thick.     At  0°  C.  what  is  its  resistance 
between  ends? 

14.  A  lead  cable  sheath  is  3  centimeters  in  outside  diameter 
and  2  millimeters  thick.     Owing  to  impurities  the  conductivity 
of  the  sheath  is  103^  that  of  pure  lead.     If  the  temperature 
coefficient  of  resistance  of  lead  is  .004  per  degree  C.,  what  will 
be  the  resistance  of  the  sheath  per  kilometer  at  25°  C.  ? 

15.  The  resistivity  of  platinum  in  centimeter  ohm  measure 
.is  .000009.     At  0°  C.  what  is  the  resistance  of  a  platinum  wire 
1  millimeter  in  diameter,  per  meter  of  length  ? 

16.  Assuming  a  temperature  coefficient  for  the  resistance  of 
platinum  of  .00247  per  degree  C.,  how  long  must  a  wire  of  No.  18 
B.S.G.  platinum  wire  be  to  have  a  resistance  of  2  ohms  at  60°  C.? 

17.  At   35°   C.  what  diameter  platinum  wire  will    have  a 
resistance  of  .01  ohm  per  foot? 

18.  The  resistivity  of  silver  in  centimeter  ohm  measure  being 
.00000149,  what  diameter  silver  wire  will  have  a  resistance  of 
10  ohms  per  kilometer? 

19.  A   circuit   is    composed    of    the  following    elements    in 
series :  5  miles  of  No.  9  B.S.G.  iron  wire ;  a  coil  of  3000  turns 
of    No.   30    B.S.G.  copper  wire,   with   an   average    length  of 
3£  inches  per   turn  ;  6   mercury  contacts,  each  5  centimeters 
long,  with  a  cross  section  of  1  square  centimeter  ;  50  inches  of 
No.  30  B.S.G.  platinum  wire.      The  copper  has  a  conductivity 


10  ELECTRICAL  PEOBLEMS 


that  of  pure  copper;  the  iron  is  impure,  and  at  this  tem- 
perature has  a  resistance  1.75  that  of  pure  iron  at  0°  C. ;  the 
platinum  and  mercury  are  pure.  The  temperature  of  the  circuit 
is  taken  at  77°F.;  if  an  electro-motive  force  of  10  volts  is 
impressed  on  the  circuit,  what  current  will  be  produced  ? 

20.  An  aluminum  wire  1  millimeter  in  diameter  and  1  meter 
long  has  a  resistance  of  .037  ohm  at  0°  C.     What  is  the  resis- 
tivity of  aluminum  in  centimeter  ohm  measure  ? 

21.  What  is  the  resistivity  of  aluminum  in  circular  mil-foot 
ohm  measure  ? 

22.  The  temperature  coefficient  of  resistance  of  aluminum  is 
.00139.      What  is   the  resistance    per  mile   of   No.   3   B.S.G. 
aluminum  wire,  at  18°  C.? 


CHAPTER    IV 

ELECTRO-MOTIVE   FORCES  AND  RESISTANCES   IN   SERIES 

GROUPING 

When  two  or  more  sources  of  electro-motive  force  are  arranged 
so  that  their  electrical  pressures  act  along  a  single  circuit,  the 
positive  terminal  of  one  being  connected  to  the  negative  terminal 
of  the  next,  either  directly  or  through  a  section  of  the  circuit, 
the  positive  terminal  of  the  last  and  the  negative  terminal  of 
the  first  being  connected  through  an  external  circuit,  the 
several  sources  of  electro-motive  force  are  said  to  be  in  series 
grouping. 

A  typical  case  is  that  of  a  zinc-carbon  battery  used  to  ring  a 
bell.  The  zinc  of  one  cell  is  connected  to  the  carbon  of  .the 
next,  the  zinc  of  the  second  cell  to  the  carbon  of  the  third, 
and  so  on,  the  free  zinc  and  carbon  terminals  of  the  battery 
being  connected  to  the  bell  terminals  by  the  usual  insulated 
wires. 

In  such  a  case  the  available  electro-motive  force  is  the 
algebraic  sum  of  all  the  component  electro-motive  forces  acting 
on  the  circuit.  When  these  all  act  in  the  same  direction  the 
resultant  is  their  arithmetical  sum. 

A  circuit  may  be  divided  into  several  sections,  the  current 
passing  to  each  one  from  the  preceding.  The  resistances  of  the 
several  sections  are  then  said  to  be  in  series  grouping.  The 
total  resistance  of  the  several  sections  arranged  in  series  is  their 
arithmetical  sum. 

Unless  otherwise  stated,  electro-motive  forces  acting  on  the 
same  circuit  are  considered  to  act  in  the  same  direction,  or 
accumulatively. 

11 


12  ELECTRICAL   PROBLEMS 

PROBLEMS 

1.  Two  electro-motive  forces,  one  of  5  volts  and  the  other 
of  27  volts,  act  in  series  on  a  circuit  of  two  sections  having 
resistances  of  3  ohms  and  5  ohms,  respectively,  arranged  in 
series.     What  current  will  flow? 

Solution.     Total  electro-motive  force  =5  +  27  volts  =  32  volts. 
Total  resistance  =  •>  -f  5  ohms  =  8  ohms. 

E 

I  =  —  —  4  amperes. 

2.  A  battery  is  composed  of  20  cells  in  series,  each  having 
an  electro-motive  force  of  1.1  volts.     Through  what  resistance, 
including  that  of  the  battery  itself,  will  it  produce  a  current  of 
f  ampere  ? 

3.  A  lighting  circuit  contains  40  lamps  in  series,  each  of 
8^  ohms  resistance.      If  the  resistance  of  the  generator  and 
line  is  8  ohms,  what  electro-motive  force  must  be  maintained 
to  supply  a  current  of  3  amperes  ? 

4.  A  lighting  circuit  3  miles  long  is  laid  in  an  underground 
duct.     Each  conductor  has  an  insulation  resistance  of  10  meg- 
ohms.    If  the  potential  difference  between  the  lines  is  1000 
volts,  what  is  the  leakage  current? 

5.  Ten  200  ohm  lamps  are  used  in  series  in  synchronizing 
two  alternators.     At  a  particular  instant  one  generator  gives  an 
electro-motive  force  of  1414  volts  in  one  direction  and  the  other 
an  electro-motive  force  of  614  volts  in  the  opposite  direction. 
What  current  flows  through  the  lamps  ? 

6.  What  is  the  current  in  the  shunt  field  of  a  generator 
having   a   terminal   voltage   of  550   if   the  field  resistance  is 
63  ohms  and  there  is  in  series  a  rheostat  with  a  resistance  of 
17  ohms? 

7.  A  storage  battery  of  30  cells  in  series  is  being  charged 
from  a  generator  whose  electro-motive  force  is   75  volts   and 
whose    resistance,    including   battery   connections,    is   .6   ohm. 
The  internal  resistance  of  the  battery  is  .03  ohm  per  cell  and 


SERIES  GROUPING  13 

its  back  electro-motive  force  is  2  volts  per  cell.     What  is  the 
charging  current? 

8.  During  the  process  of  charging  the  battery  of  Problem  7 
the  back  electro-motive  force  rises  to  2.1  volts  per  cell.     What 
generator  electro-motive  force  is  needed  to  keep  the  charging 
current  constant? 

9.  Keeping  the  electro-motive    force   of   the  generator   of 
Problem  7  constant,  what  will  be  the  charging  current  when 
the  back  electro-motive  force  has  risen  to  2.1  volts  per  cell? 

10.  A  battery  of  10  cells  in  series  acts  through  an  external 
circuit  of  10  ohms  resistance.     On  open  circuit  each  cell  has  an 
electro-motive  force  of  1.5  volts  and  an  internal  resistance  of 
1  ohm.     After  a  period  of  activity  the  effect  of  polarization 
is  to  increase  the  resistance  .2  ohm  per  cell  and  to  diminish  the 
effective  electro-motive  force  .4  volt  per  cell.     At  this  stage, 
by  what  per  cent  is  the  current  reduced  from  its  value  at  the 
time  of  closing  the  circuit? 

11.  There   are  impressed  on  a  circuit  three  electro-motive 
forces  of  3,  4,  and  5  volts,  respectively.     The  sources  of  electro- 
motive forces  have  internal  resistances  as  follows :  that  of  3  volts 
4  ohms,  that  of  4  volts  5  ohms,  and  that  of  5  volts  6  ohms. 
The  resistance  of  the  external  circuit  is  15  ohms.     What  is  the 
current  ? 

12.  In  Problem  11  what  is  the  current  if  the  3  volts  electro- 
motive force  acts  in  opposition  to  the  other  two  ? 

13.  In  Problem  11  what  is  the  current  if  the  4  volts  electro- 
motive force  acts  in  opposition  to  the  other  two  ? 

14.  In  Problem  11  what  is  the  current  if  the  5  volts  electro- 
motive force  acts  in  opposition  to  the  other  two  ? 

15.  What  is  the  current  in  Problem  11  if  the  source  of  3  volts 
electro-motive  force  is  removed? 

16.  What  is  the  current  in  Problem  11  if  the  source  of  4  volts 
electro-motive  force  is  removed? 

17.  What  is  the  current  in  Problem  11  if  the  source  of  5  volts 
electro-motive  force  is  removed  ? 


14  ELECTRICAL   PROBLEMS 

18.  What  is  the  current  in  Problem  11  if  the  external  resist- 
ance is  removed  ? 

19.  The  difference  of  potential  between  the  carbons  of  an  arc 
lamp  is  40  volts.     It  is  assumed  that  the  arc  has  a  back  electro- 
motive force  of  35  volts.     If  a  current  of  10  amperes  flows, 
what  is  the  resistance  of  the  arc  ? 

20.  A  motor  driven  from  a  battery  of  60  volts   potential 
takes  a  current  of  5  amperes.     If  the  resistance  of  the  battery 
is  2  ohms  and  that  of  the  motor  and  connections  1.5  ohms, 
what  is  the  back  electro-motive  force  of  the  motor  ? 

21.  A  3  phase  dynamo  has  4  collecting  rings,  A,  B,  C,  D. 
One  end  of  each  coil  is  connected  to  a  common  ring  and  the 
other  ends  are  connected,  each  to  a  separate  ring.     A  current 
is  passed  successively  from  each  ring  to  each  of  the  others. 
An   ampere-meter    is    placed  in   the  circuit  and  a    voltmeter 
connected  to  the  rings  between  which  the  current  passes  with 
the  following  readings : 

From  To  Volts  Amperes 

A  B  i«  9.2 

A  C  §»  8.7 

A  D  f»  8.8 

B  C  ™  9.3 

B  D  if  9.2 

C  D  |§  8.7 

Find  the  resistance  between  each  pair  of  rings  and  state  which 
ring  is  connected  to  the  common  junction. 

22.  Ten  cells  in  series  have  each  an  internal  resistance  of  1  ohm 
and  give  an  electro-motive  force  of  1.1  volts.    There  is  an  external 
copper  circuit  having,  at  0°  C.,  a  resistance  of  If  ohms.    If  the 
copper  is  placed  in  boiling  water,  what  current  will  flow  ? 

23.  The  same  battery  is  used  as  in  Problem  22.     The  exter- 
nal resistance  at  50°  C.  is  10  ohms  copper  and  10  ohms  iron. 
The  temperature  coefficient  for  the  resistance  of  iron  is  .0054 
per  degree  C.     If  the  copper  is  cooled  to  10°  C.  and  the  iron 
is  heated  to  90°  C.,  what  current  will  flow? 


SERIES   GROUPING  15 

24.  With  the  conditions  the  same  as  in  Problem  23,  except 
that  the  copper  is  heated  to   90°  C.  and  the  iron  cooled   to 
10°  C.,  what  current  will  flow? 

25.  With  conditions  the  same  as  in  Problem  23,  except  that 
the  copper  is  heated  to  80°  C.  and  the  iron  to  210°  C.,  what 
current  will  flow? 

26.  Suppose  the  heating  in  Problem  25  to  have  been  caused 
by  the  flow  of  the  current.     In  the  meantime  polarization  has 
diminished  the  effective  electro-motive  force  .3  volt  per  cell,  and 
has  increased  the  internal  resistance  .4  ohm  per  cell.     Under 
these  conditions,  what  would  be  the  current  in  the  circuit  ? 

27.  An  electro-motive  force  of  100  volts  is  maintained  at  the 
terminals  of  a  circuit  which  at  0°  C.  consists  of  40  ohms  copper 
and  60  ohms  iron  in  series.     Plot  a  curve  of  current  as  the 
temperature  of  the  circuit  rises  to  350°  C. 


CHAPTER   V 

ELECTRO-MOTIVE  FORCES  AND  RESISTANCES  IN  PARALLEL 

GROUPING 

The  ability  of  a  conductor  to  transmit  a  uniform  current 
without  excessive  heating  varies  inversely  with  its  resistance  or 
directly  with  the  reciprocal  of  its  resistance,  its  conductance. 
The  idea  of  conductance  is  especially  useful  in  dealing  with 
multiple  or  branched  circuits  where  the  joint  conductance  of 
the  parallel  branches  is  the  sum  of  the  conductances  of  the 
individual  branches.  It  follows,  therefore,  that  the  joint  resist- 
ance  of  two  or  more  conductors  in  multiple  is  the  reciprocal 
of  the  sum  of  the  reciprocals  of  their  separate  resistances. 
Hence  the  joint  resistance  of  a  branched  circuit  is  less  than 
the  resistance  on  any  single  branch. 

The  law  of  Ohm  may  be  expressed  as  follows  : 


where  K  is  the  conductance,  the  unit  of  which  has  been  called 
the  mho. 

When  two  or  more  equal  electro-motive  forces  are  so  arranged 
that  they  act  all  in  the  same  direction  along  a  given  circuit, 
all  the  positive  terminals  being  connected  at  one  point  of  the 
circuit  and  all  the  negative  terminals  being  joined  at  another 
point  of  the  circuit,  these  common  junctions  being  the  terminals 
of  the  group,  they  are  said  to  be  connected  in  parallel  or 
multiple  grouping.  The  electro-motive  force  of  the  group 
as  a  whole  is  the  same  as  that  of  any  one  of  its  individual 
elements. 

16 


PARALLEL  GROUPING  17 

(.?. 

PROBLEMS 

1.  A  divided  circuit  has  two  branches  of  1  ohm  and  J  ohm, 
respectively.     What    is    the    joint    conductance    of   the    two 
branches  ?     What  is  the  joint  resistance  ? 

2.  A  divided  circuit  has  resistances  in  the  several  branches 
of  1,  2,  4,  5,  and  10  ohms,  respectively.    What  is  the  conductance 
of  the  combination  ?     What  is  the  resistance  ? 

3.  To  produce  a  current  of  8  amperes,  what  electro-motive 
force  must  be  impressed  on  a  circuit  containing  the  following 
elements  in  series :  a  coil  of  6  ohms  in  multiple  with  one  of 
2  ohms,  a  line  of  10  ohms,  a  group  of  10  incandescent  lamps  in 
multiple,  each  of  200  ohms,  a  connecting  wire  of  4  ohms? 

4.  A  circuit  has  two  branches,  one  of  2  ohms  with  a  current 
of  6  amperes,  the  other  of  15  ohms.    What  current  flows  in  the 
second  branch  and  what  difference  of  potential  is  maintained 
between  the  terminals  of  the  circuit? 

5.  A  circuit  has  three  branches  of  12,  3,  and  6  ohms,  respec- 
tively.   If  4  amperes  flow  in  the  circuit  containing  3  ohms,  what 
current  will  flow  in  each  of  the  others  ? 

6.  What  is  the  resistance  of  a  circuit  of  A  branches,  each 
containing  B  ohms  ? 

7.  Twenty  cells,  each  of   2  volts  electro-motive  force  and 
1  ohm  resistance,  are  connected  in  parallel  to  an  external  circuit  of 
1  ohm  resistance.   What  current  will  flow  in  the  external  circuit  ? 

8.  A  battery  cell  with  an  electro-motive  force  of  2  volts  and 
an  internal  resistance  of  2  ohms  acts  through  an  external  circuit 
of  2  ohms  resistance.     By  what  per  cent  will  the  current  be 
increased  by  putting  another  similar  cell  in  multiple  with  the 
first? 

9.  In  Problem  8  what  would  have  been  the  percentage  increase 
in  the  current  by  putting  the  second  cell  in  series  with  the  first  ? 

10.  How  many  cells  like  that  of  Problem  8  must  be  put  in 
multiple  to  produce  a  current  of  1  ampere  in  an  external  circuit 
of  If  ohms  resistance? 


18  ELECTRICAL  PROBLEMS 

11.  A  certain  cell  has  an  electro-motive  force  of  1.1   volts 
and  a  resistance  of  1   ohm.     Twenty  such  cells  in  series  act 
through  a  circuit  of  24  ohms.    What  is  the  current? 

12.  The  battery  of  Problem  11  is  arranged  with  the  cells  in 
2  groups,  each  of  10  in  series,  the  2  groups  being  in  multiple. 
What  is  the  current? 

13.  The  battery  of  Problem  11  is  arranged  with  the  cells  in 

4  groups,  each  of  5  in  series,  the  4  groups  being  in  multiple. 
What  is  the  current? 

14.  The  battery  of  Problem  11  is  arranged  with  the  cells  in 

5  groups,  each  of  4  in  series,  the  5  groups  being  in  multiple. 
What  is  the  current  ? 

15.  The  battery  of  Problem  11  is  arranged  with  the  cells  in 
10  groups,  each  of  2  in  series,  the  10  groups  being  in  multiple. 
What  is  the  current  ? 

16.  The  battery  of  Problem  11  is  arranged  with  all  the  cells 
in  multiple.     What  is  the  current? 

17.  With  an   external   resistance   of  20   ohms,  what  is  the 
greatest  possible  current  with  the  battery  of  Problem  11,  and 
by  what  arrangement  of  cells  will  it  be  produced  ? 

18.  With   an    external   resistance    of   5   ohms,  what   is    the 
greatest  possible  current  with  the  battery  of  Problem  11,  and  by 
what  arrangement  of  cells  will  it  be  produced? 

19.  With  an  external  resistance  of  1.25   ohms,  what  is  the 
greatest  possible  current  with  the  battery  of  Problem  11,  and  by 
what  arrangement  of  cells  will  it  be  produced  ? 

20.  With  an  external  resistance  of  .8  ohm,  what  is  the  greatest 
possible  current  with  the  battery  of  Problem  11,  and  by  what 
arrangement  of  cells  will  it  be  produced  ? 

21.  With  an  external  resistance  of  .2  ohm,  what  is  the  greatest 
possible  current  with  the  battery  of  Problem  11,  and  by  what 
arrangement  of  cells  will  it  be  produced  ? 

22.  With   an   external   resistance    of   .05   ohm,  what  is   the 
greatest  possible  current  with  the  battery  of  Problem  11,  and 
by  what  arrangement  of  cells  will  it  be  produced  ? 


PARALLEL   GROUPING  19 

The  student  will  note  that  in  Problems  17-22  the  arrange- 
ment of  cells  giving  the  maximum  current  is  that  in  which  the 
internal  and  external  resistances  are  as  nearly  as  possible  equal. 
In  general,  with  batteries  or  other  sources  of  electro-motive  force, 
when  all  the  sources  of  electro-motive  force  are  similar  as  to 
voltage  and  resistance,  the  arrangement  for  maximum  current 
output  is  that  which  makes  the  ratio  of  the  internal  and  external 
resistances  as  nearly  as  possible  equal  to  unity. 

23.  Each  of  two  dynamos  has  an  electro-motive  force  of  1000 
volts,  and  a  resistance  of  2  ohms  between  terminals.     With  an 
external  resistance  of  40  ohms,  what  current  will  be  supplied 
with  the  dynamos  in  series,  and  what  with  them  in  parallel  ? 

24.  If  the  resistance  between  opposite  faces  of  an  inch  cube 
of  a  battery  solution  is  10  ohms,  what  is  the  internal  resistance 
of  a  battery  having  two  plates,  each  of  10  square  inches  area, 
1   inch  apart? 

25.  A  battery  cell  consists  of  10  positive  and  11  negative 
plates,    15    inches   by  20  inches  in  size,   arranged    alternately 
positive  and  negative,  and  separated  J  inch.     The  battery  solu- 
tion has  a  resistivity  of  300  in  inch  ohm  measure.    What  is  the 
internal  resistance  of  the  cell  ? 

26.  A  lighting  circuit  5  miles  long  is  laid  in  an  underground 
duct,  each  wire  having  an  insulation  resistance  of  four  megohms 
per  mile.     If  the  potential  difference  between  the  wires  is  2000 
volts,  what  is  the  leakage  current  ? 

27.  It  is  desired  to  shunt  a  galvanometer  of  10  ohms  resistance 
so  that  only  Ijfo  of  the  current  shall  pass  through  it.     What 
must  be  the  resistance  of  the  shunt? 

28.  If  in  Problem  27  the  remainder  of  the  circuit  has  a  resist- 
ance of   10   ohms,  how   and  by  what  per  cent  will  the  main 
current  be  affected  by  the  insertion  of  the  above  shunt  ?    What 
will  be  the  effect  on  the  galvanometer  current? 

29.  What  resistance  in  series  with  the  above-shunted  galva- 
nometer will  compensate   for  the  insertion  of  the  shunt  so  as 
not  to  alter  the  main  current  ? 


20  ELECTRICAL  PROBLEMS 

30.  Calculate  a  y1^-,  a  y-g-g-,  and  a  yg-^-g-  set  of  shunts  for  a  3000 
ohm  galvanometer. 

31.  A   5000   ohm  galvanometer  is  connected  to  a  thermo- 
electric pile  of  3  ohms  resistance.     How  and  by  what  per  cent 
will    the  insertion   of  a  y^oo"  galvanometer  shunt  affect   the 
current  in  the  galvanometer? 

32.  A  galvanometer  of  4000  ohms  resistance  is  provided  with 
a  set  of  shunts  and  compensating  resistances  operated  in  the 
following  manner :   inserting  the  shunt  plug  in  the  first  hole 
connects  a  y-^-g-  shunt,  sv  across  the  terminals  of  the  galvanom- 
eter,  and  at  the  same  time  inserts  a  compensating  resistance 
consisting  of  coils  rv  r2,  and  r3,  connected  in  series ;  putting  the 
plug  in  the  second  hole  connects  a  yj-g-  shunt,  s2,  across  the 
galvanometer  in  series  with  rv  r2  in  series  with  rs  acting  as 
the  compensating  resistance.     When  the  plug   is  in  the  third 
hole  a  J-g-  shunt,  s3,  is  connected  across  the  galvanometer  in 
series  with    rl    and  r2,    r3  being  the   compensating  resistance. 
Calculate  the  values  of  sv  s2,  «3,  rr  r2,  and  ry 


.  > 


. 

CHAPTER   VI 


The  difference  of  potential  or  the  drop  in  voltage,  frequently 
termed  simply  drop,  between  any  two  points  in  a  circuit  in 
which  a  constant  current  is  flowing,  is  numerically  equal  to 
the  product  of  the  current  flowing  in,  and  the  resistance  of,  the 
conductor  between  the  given  points. 

A  little  consideration  will  show  that  the  sum  of  the  separate 
drops  in  the  various  sections  into  which  a  circuit  may  be  divided 
must  equal  the  algebraic  sum  of  all  the  electro-motive  forces  in 
the  circuit  ;  evidently,  also,  in  the  case  of  a  divided  circuit  the 
drop  is  the  same  in  each  of  several  parallel  branches. 


PROBLEMS 

1.  If  a  current  of  1  ampere  flows  in  a  circuit,  what  is  the 
drop  in  a  section  having  a  resistance  of  1  ohm? 

Solution.     Drop  =  1R  =1x1  =  1  volt. 

2.  A  trolley  wire  of  No.  0  B.S.G.  has  a  resistance  of  .519  ohm 
per  mile.     What  is  the  copper  drop  between  the  station  and  a 
car  taking  20  amperes,  2  miles  out  on  the  line  ? 

NOTE.  —  In  the  following  problems  trolley  wire  will  be  assumed  No.  0. 

3.  A  power  station  maintains  a  difference  of  potential  of  550 
volts  between  trolley  and  ground  at  the  station.    The  resistance 
of  the  rail  return  is  .04  ohm  per  mile.     If  there  is  only  one  car 
on  the  line,  how  far  from  the  station  must  it  be  to  have  the 
pressure  fall  to  500  volts  with  a  current  of  35  amperes? 

21 


22  ELECT1I1CAL    PROBLEMS 

4.  Seven  cars  a  mile  apart  and  each  taking  15  amperes  an 
on  a  trolley  line  6  miles  long.     Beginning  at  the  station,  find  tlw 
drop  to  each  car,  assuming  .04  ohm  per  mile  for  the  rail  return 

NOTE.  —  This  problem  illustrates  the  importance  of  supplying  trolley 
lines  with  large  supplementary  conductors,  called  feeders,  and  of  frequently 
connecting  these  feeders  to  the  trolley  wire.  By  this  means  the  resistance 
of  the  line  is  kept  down  and  excessive  drops  prevented. 

5.  If   the   station   in  Problem   4   maintains   a   potential   o 
600  volts,  what  is  the  voltage  at  each  of  the  seven  cars? 

6.  A  circuit  consists  of  8  ohms  resistance  in  generator  am' 
line,  and  of  40  lamps  in  series,  each  of  8£  ohms.  With  a  curren 
of  3  amperes,  what  will  be  the  drop  in  the  generator  and  line 
and  what  the  fall  in  voltage  at  each  lamp  ? 

7.  An  arc  light  dynamo  of  30  ohms  resistance  supplies  a  cur 
rent  of  6.8  amperes  through  15  miles  of  line  wire  of  No.  6  B.S.G 
to  a  series  of  45  arcs,  each  adjusted  to  47  volts.    Find  the  electro 
motive  force  of  the  dynamo,  the  difference  of  potential  at  it: 
terminals,  and  the  drop  on  the  line  at  60°  F.  and  at  0°  F. 

8.  A   lighting  circuit  consists   of  10   groups  of   lamps   ir 
multiple  between  the  line  wires,   the   groups    being  100  fee 
apart,   and    the    nearest    group    500   feet   from    the   generator 
Each  group  of  lamps  takes  5  amperes,  and  the  resistance  o 
the  line  is  .102  ohm  per  1000  feet.     What  is  the  drop  from  th( 
generator  to  each  group  ? 

9.  Instead  of  the  arrangement  described  in  Problem  8,  one 
of  the   two   line   wires   runs    directly   to   the  group   of  lamp; 
farthest  from  the  generator,  and  then  back  to  the  group  nearest 
connecting  with  each  group  on  the  return.     All  other  condition; 
are  as  in  Problem  8.     What  is  the  drop  to  each  group? 

10.  Plot  curves  showing  the  conditions  in  Problems  8  and  9 
using  for  abscissas  the  distances  of  the  various  groups  from  th< 
generator,  and  for  ordinates  the  voltages  at  the  groups.    Assmm 
in  each  case  that  the  voltage  of  the  generator  is  115  volts. 

11.  An  electric  railway  6  miles  long  is  equipped  with  No.  ( 
trolley  and  4  miles  of  feeder  of  No.  00,  with  a  resistance  o 


DKOP  23 

.412  ohm  per  mile,  connected  to  the  trolley  every  half  mile. 
At  a  certain  instant  of  time  4  cars  are  running  as  follows  : 
No.  1  is  1  mile  from  the  station,  taking  10  amperes ;  No.  2  is 
2  J  miles  from  the  station,  taking  40  amperes ;  No.  3  is  4£  miles 
from  the  station,  taking  25  amperes ;  No.  4  is  5i  miles  from 
the  station,  taking  30  amperes.  Neglecting  the  rail  resistance, 
what  is  the  voltage  at  each  car  if  the  station  voltage  is  550  ? 

12.  Assuming  a  70  pound  rail  with  a  track  resistance  of  .03 
ohm  per  mile,  what  must  be  the  station  voltage  in  Problem  11 
to  give  500  volts  at  the  last  car? 

13.  An  electric  railway  5  miles  long  has  two  feeders ;    the 
first  is  No.  00  and  extends  4  miles  from  the  station,  being  tied 
to  the  trolley  every  half  mile ;  the  other  is  No.  0000,  with  a 
resistance  of  .259  ohm  per  mile,  is  2  miles  long,  and  is  connected 
to  the  trolley  at  its  farther  end.     The    cars    are    situated    as 
follows :  No.  1  is  starting  from  the  station,  taking  40  amperes ; 
No.  2  is  li  miles  out,  taking  30  amperes ;  No.  3  is  3  miles 
out,   taking   10   amperes  ;    No.   4   is   4^  miles  out,  taking   50 
amperes.     Neglecting  the  track  resistance  and  assuming  that 
the  voltage  at  the  car  3  miles  out  is  500,  find  the  voltage  at  the 
station  and  at  each  of  the  other  cars. 

SUGGESTION.  —  In  solving  this  problem  we  will  first  determine  the 
distribution  of  current  in  the  three  conductors.  This  we  may  do  by 
remembering  that  the  drop  in  the  2  miles  of  No.  0000  feeder  must  be  the 
same  as  that  in  the  same  length  of  trolley  and  No.  00  feeder.  Hence  call 
the  current  in  the  first  1£  miles  of  trolley  and  No.  00  feeder  X,  and  the 
current  in  the  No.  0000  feeder  F;  then  X  +  F  =  90,  and  1.5  x  .2297  X  + 
.5  x  .2297  (X  -  30)  =  2  x  .259  F. 

14.  Assuming  the  track  resistance  to  be  .025  ohm  per  mile, 
and  keeping  the  station  voltage  in  Problem  13  unaltered,  what 
will  be  the  voltage  at  each  car? 

15.  A  railway  7  miles  long  has  a  trolley,  a  No.  00  feeder 
6  miles  long  tied  to  the  trolley  at  every  half  mile,  and  a  No.  0000 
feeder  4  miles  long  connected  to  the  overhead  system  at  each 
mile ;  7  cars  a  mile  apart  are  on  the  line,  the  first  being  a  mile 


24  ELECTRICAL  PROBLEMS 

from  the  station,  and  each  taking  25  amperes.     Allowing  .0212 
ohm  per  mile  for  track  return,  what  is  the  drop  to  each  car? 

16.  A  trolley  road  is  arranged  in  the  form  of  a  square  with 
sides  2  miles  long,  and  the  station  in  the  middle  of  one  side. 
From  the  station  a  No.  0000  feeder  runs  directly  across  the 
square  to  the  middle  of  the  opposite  side,  where  it  is  connected 
to  a  No.  00  feeder  extending  .5  mile  each  way.     From  the  station 
a  No.  00  feeder  extends  2.5  miles  in  each  direction  around  the 
square.     All  No.  00  feeders  are  connected  to  the  trolley  at  every 
half  mile.     A  car  at  the  middle  of  each  side  of  the  square  takes 
50  amperes,  and  one  at  each  corner  takes  20  amperes.     Draw  a 
diagram  of  the  road  and  determine  the  current  in  each  conduc- 
tor and  the  voltage  at  each  car,  assuming  525  volts  at  the  station, 
but  neglecting  the  resistance  of  the  rail  return. 

17.  Allowing    .0212    ohm   per   mile  for  the   rail  return  in 
Problem  16,  by  what  amount  will  the  voltage  at  each  car  be 
diminished  by  the  drop  in  the  return  circuit? 

18.  An  electric  road  is  in  the  form  of  a  rectangle  with  sides 
of  2  and  4  miles,  and  the  station  is  in  the  middle  of  one  of  the 
shorter  sides.     From  the  station  a  feeder  with  a  resistance  of 
.1  ohm  per  mile  extends  across  the  rectangle  to  the  middle  of 
the  opposite  side.     Connecting  the  middles  of  the  longer  sides 
and  the  above-described  feeder  is  a  cross  feeder  of  .2  ohm  per 
mile.     The  combined  resistance  of  trolley  and  feeder  extending 
either  way  from  the  station  to  the  middle  of  the  longer  sides  is 
.2  ohm  per  mile,  from  these  points  to  farther  corners  it  is  .3  ohm 
per  mile,  and  from  the  corners  to  the  middle  of  the  side  opposite 
the  station  it  is  .4  ohm  per  mile ;  50  amperes  are  required  at 
the  middle  of  each  side  and  at  each  corner  of  the  rectangle,  and 
20  amperes  at  each  point  midway  between.     Determine  the  dis- 
tribution of  current  and  the  drop  to  each  of  the  above-described 
points,  neglecting  the  resistance  of  the  rail  return. 

19.  Assuming  550  volts  at  the  station  and  .02  ohm  per  mile 
for  the  resistance  of  the  rail  return,  determine  the  voltage  at 
each  of  the  required  points  on  the  line,  in  Problem  18. 


CHAPTER    VII 
POWER  AND  EFFICIENCY 

The  power,  activity,  or  time  rate  of  expenditure  of  energy  in 
a  circuit  is  numerically  equal  to  the  product  of  the  current  and 
electro-motive^  force  in  the  circuit.  This  may  be  expressed  by 
the  equation  P  =  EL 

Thus  in  practical  units : 

watts  —  joules  per  second  —  volts  x  amperes. 
1  horse  power  —  746  watts. 

Furthermore  the  power  absorbed  in  any  portion  of  a  cir- 
cuit is  equal  to  the  product  of  the  current  and  difference  of 
potential  between  the  extremities  of  that  portion  of  the  circuit, 
p  =  el. 

Since  e,  the  difference  of  potential  between  two  given  points 
of  a  circuit,  is  equal  to  the  product  of  the  current  and  resistance 
between  the  given  points,  e  =  Ir,  it  follows  that  p  =  I2r  for  any 
portion  of  a  circuit. 

PROBLEMS 

1.  A  generator  delivers  a  current  of  100  amperes  at  a  pres- 
sure of  100  volts.     What  power  does  it  supply  in  kilowatts? 
How  many  horse  power  ? 

2.  A  current  of  1.5  amperes  flows  through  a  circuit  of  2  ohms 
resistance.     How  many  watts  are  absorbed  ? 

3.  How  many  watts  are  expended  in  a  110  volt  16  candle 
power  lamp  requiring  a  current  of  .5  ampere  ?     How  many  watts 
per  candle  ? 

25 


26  ELECTRICAL   PROBLEMS 

4.  The  generator  in   Problem  1   has  a  shunt  wound  field 
with  a  resistance  of  50  ohms.     How  many  watts  are  absorbed  in 
the  field  coils  ? 

5.  The    armature    of  above   generator  has   a  resistance  of 
.03  ohm.     What  power  is  lost  in  the  armature  conductors? 

6.  Assuming  all  other  losses  to  be  1  kilowatt,  how  many 
horse  power  will  be  required  to  drive  above  generator  at  full 
load  of  10  kilowatts? 

7.  An  engine  supplies  145  horse  power  to  a  generator  deliv- 
ering 181.82  amperes  at  an  electro-motive  force  of  550  volts. 
What  is  the  commercial  efficiency  of  the  generator? 

Solution.  145  H.P.  =  145  x  746  =  108,170  watts,  input, 

181.82  x  550  =  100,000  watts,  output. 

^^=.9245  =  92.45%  efficiency. 

108,170 

8.  What  is  the  commercial  efficiency  of  the  generator  of 
Problems  1,  4,  5,  and  6  ? 

9.  A  battery  is  formed  of  2  cells,  each  with  2  volts  electro- 
motive force  and  1  ohm  resistance.     The  cells  are  connected  in 
series  and  supply  current  to  a  circuit  of  2  ohms  resistance.     How 
much  power  is  absorbed-in  the  battery?  how  much  in  the  exter- 
nal circuit,  and  what  is  the  efficiency  ? 

NOTE.  —  In  the  case  of  a  battery  the  efficiency  is  the  useful  power,  that 
is,  the  power  absorbed  in  the  external  circuit,  divided  by  the  total  power 
generated. 

10.  The  cells  of  Problem  9  are  connected  in  parallel.     What 
power  is  absorbed  in  the  battery  ?  what  in  the  external  circuit, 
and  what  is  the  efficiency  ? 

11.  A  battery  has  24  cells  like  those  of  Problems  9  and  10. 
They  may  be  connected  in  8  different    ways :    all   in   series  ; 
with  12  pairs  in  series,  each  pair  in  parallel;  with  8  groups  of  3 
in  series,  each  group  in  parallel ;   with  6  groups  of  4  in  series, 
each  group  in  parallel ;  with  4  groups  of  6  in  series,  each  group 


POWER  AND  EFFICIENCY  27 

in  parallel ;  with  3  groups  of  8  in  series,  each  group  in  parallel ; 
with  2  groups  of  1 2  in  series,  each  group  in  parallel ;  all  in 
parallel.  If  the  external  circuit  has  4  ohms  resistance,  what 
power  is  developed  by  the  battery  in  each  case  ? 

12.  Under  the  conditions  of  Problem  11,  what  power  is  avail- 
able in  the  external  circuit  in  each  case  ? 

13.  Under  the  conditions  of  Problem  11,  what  is  the  efficiency 
of  the  system  in  each  case  ? 

14.  Under  the  conditions  of  Problem  11,  what  per  cent  of  the 
total  resistance  is  in  the  external  circuit  in  each  case  ? 

15.  Plot  a  set  of  5  curves  for  the  above  battery,  using  for 
abscissas  the  per  cent  of  the  total  resistance  included  in  the 
external  circuit  and  for  ordinates  the  current,  the  total  power 
generated,  the  useful  power,  the  power  absorbed  in  the  battery, 
and  the  efficiency. 

16.  A 100  kilowatt  generator,  giving  a  terminal  electro-motive 
force  of  125  volts,  has  an  armature  resistance  of  .005  ohm  and 
a  brush  contact  resistance  of  .00125  ohm  on  each  side.     The 
total  loss  by  friction  is  2100  watts,  constant  at  all  loads.     The 
armature  core  loss  is  1400  watts  at  25%  load,  1500  watts  at  50% 
load,  1700  watts   at  75%  load,   2000  watts  at  full  load,  and 
2400  watts  at  125%  load.     The  power  absorbed  by  the  field  is 
546,  590,  634,  678,  and  722  watts  at  25%,  50%,  75%,  100%, 
and  125%  of  full  load,  respectively.     What  is  the  total  loss  at 
each  of  the  above  percentages  of  full  load  ? 

17.  What  is  the  commercial  efficiency  of  the  generator  of 
Problem  16  at  each  of  the  specified  percentages  of  load? 

18.  A  storage  battery  consists  of  60  cells  with  an  internal 
resistance  of  .003  ohm  each.    In  charging,  the  cells  are  connected 
in  2  sets  in  multiple,  each  of  30  in  series.     The  charging  cur- 
rent is  delivered  by  a  generator  having  .15  ohm  resistance, 
through  a  line  of  .02  ohm  resistance.     At  the  time  of  begin- 
ning the  charge  the  battery  gives  a  back  electro-motive  force  of 
1.95  volts  per  cell.    What  electro-motive  force  must  be  induced 
in  the  generator  to  give  a  charging  current  of  50  amperes? 


28  ELECTRICAL  PROBLEMS 

19.  Assuming  the  electrical  efficiency  of  such  a  system  as 
that  of  Problem  18  to  be  the  percentage  of  the  generated  electro- 
motive force  used  to   overcome  the  back    electro-motive  force 
of  the  battery,  what  is  the  electrical  efficiency  of  the  system  of 
Problem  18? 

20.  If  all  the  cells  of  the  battery  of  Problem  18  are  connected 
in  series,  what  electro-motive  force  must  be  generated  to  keep 
the  charging  current  25  amperes  per  cell  ? 

21.  What  is  the  electrical  efficiency  of  the  system  of  Prob- 
lem 20  ? 

22.  When  fully  charged  the  cells   of  Problem  18   have   a 
back  electro-motive  force  of  2.2  volts  each.     If  connected  as 
in  Problem  18,  what  electro-motive  force  must  be  generated 
at  the   end    of   the    charging  to  keep  the   current  up   to  its 
original  value?     What 'is  the  electrical  efficiency  of  the  system 
at  this  time  ? 

23.  When  the  battery  is  fully  charged,  what  electro-motive 
force  must  be  generated  to  keep  the  charging  current  at  its 
original  value  with  the  connection  of  Problem  20  ? 

24.  An  arc  light  dynamo  generates  a  current  of  9.6  amperes. 
The  resistance  of  the  machine  is  19  ohms;   there  are  12  miles 
of  No.  6  B.S.G.  line  wire  at  a  temperature  of  60°  F.  and  40 
arcs,  each  taking  48  volts.     What  per  cent  of  the  total  elec- 
trical  power   is    used   in  the   dynamo,  what  per  cent  on  the 
line,  and  what  per  cent  is  utilized  in  the  lamps  ? 

25.  What  is   the   electrical  efficiency  of  a  6.8   ampere  arc 
dynamo  maintaining  a  terminal  potential  difference   of  2700 
volts  and  having  an  internal  resistance  of  41.25  ohms  ? 

26.  Assuming  an  efficiency  of  50%  for  the  whole  arrange- 
ment, what  current  will  be  used  by  a  250  volt  electric  hoist 
when  raising  2500  pounds  200  feet  per  minute? 

27.  Assuming  an  average  efficiency  of  30%,  what  current 
will  be  required  by  a  500  volt  railway  motor  to  impart  a  velocity 
of  10  miles  per  hour  in  20  seconds,  to  an  8  ton  car  on  a  level 
track? 


POWER  AND  EFFICIENCY  29 

28.  Assuming  that  70%   of  the  electrical  input  is  usefully 
expended,  what  current  will  be  required  by  a  500  volt  railway 
motor  in  propelling  a  6  ton  car  up  an  8  %  grade  at  a  speed  of 
8  miles  per  hour? 

29.  A  copper  wire  of  18.7  mils  diameter  is  heated  to  redness 
by  a  current  of  19  amperes,  at  which  time  there  is  a  difference 
of  potential  of  4£  volts  between  points  on  the  wire  24.85  inches 
apart.     How  many  horse  power  will  be  expended  in  heating  a 
hollow  copper  cylinder  4  inches  in  diameter  and  4  feet  long  to 
the  same  temperature  ? 


CHAPTER    VIII 
THE   MAGNETIC  FIELD   DUE   TO   A  CURRENT 

It  is  shown  in  electrical  treatises  that  the  magnetic  force  at  a 
point,  due  to  a  current  in  a  long  rectilinear  conductor,  varies 
directly  as  the  current  and  inversely  as  the  distance  of  the  point 
under  consideration  from  the  conductor. 

Expressed  in  practical  units,  we  have 


where  If  is  the  strength  of  the  field  in  gausses,  /  the  current  in 
amperes,  and  r  the  perpendicular  distance  in  centimeters  of  the 
point  from  the  conductor.  In  a  vacuum,  and  practically  in  air 
and  other  feebly  magnetic  media,  a  magnetic  force  H  produces 
a  numerically  equal  flux  density  B,  whose  unit  is  also  called  a 
gauss.  In  magnetic  media  the  magnetic  flux  per  square  centi- 
meter, flux  density,  or  magnetic  induction  B  is  equal  to  the 
magnetic  force  H  multiplied  by  ^,  the  magnetic  permeability  of 
the  medium. 


p  is  unity  for  a  vacuum,  slightly  greater  than  unity  for  air 
and  most  other  substances,  and  possesses  a  considerable  value 
only  in  iron,  nickel,  cobalt,  and  their  alloys  or  compounds. 

.27 

In  general,  then,  we  may  say  that  B  =  -  »  but  when  the 

space  about  the  conductor  is  occupied  by  iron 


r 

30 


THE  MAGNETIC  FIELD  DUE  TO  A  CURBEOT       31 


PROBLEMS 

1.  What  is  the  strength  of  the  magnetic  field  10  centimeters 
from  a  wire  in  which  there  is  a  current  of  20  amperes  ? 

2.  What  is  the  flux  density  2.5  centimeters  from  a  straight 
wire  carrying  a  current  of  6  amperes  ? 

3.  What  is  the  field  3  feet  from  a  railway  feeder  in  which  a 
current  of  1000  amperes  is  flowing? 

4.  A  point  and  2  parallel  conductors,  2  centimeters  apart,  are 
in  a  common  plane,  the  point  being  6  centimeters  from  the 
nearer  of  the  two  conductors.     A  current  of  6  amperes  flows 
in  the  nearer  conductor,  and  an  oppositely  directed  current  of 
4  amperes  in  the  other.     What  is  the  resultant  magnetic  force 
at  the  given  point? 

NOTE.  —  At  any  point  the  direction  of  the  magnetic  force  due  to  a 
current  is  perpendicular  to  the  conductor  and  to  the  normal  from  the  con- 
ductor to  the  given  point.  Moreover  the  directions  of  the  current  and  of 
the  resulting  magnetic  force  have  the  same  relation  as  the  forward  and 
rotational  motions  of  a  right-handed  screw.  Thus,  if  an  observer  looks 
along  a  current  in  the  direction  of  flow,  the  magnetic  force  is  concentric 
with  the  conductor  and  in  the  direction  of  motion  of  the  hands  of  a  watch. 

5.  Three  parallel  wires  are  each  1  foot  from  the  other  two ; 
in  two  of  the  wires  there  is  a  current  of  100  amperes,  each  in 
the  same  direction.     What  is  the  field  in  the  space  occupied  by 
the  third  wire  ? 

SUGGESTION.  —  The  magnetic  forces  due  to  the  two  currents  will  differ 
60°  in  direction.  Their  resultant  will  be  found  by  the  same  process  that 
would  be  employed  in  finding  the  resultant  of  two  mechanical  forces. 

6.  With  the  arrangement  of  Problem  5,  what  is  the  field  at 
the  third  wire  with  the  currents  opposite  in  direction? 

7.  If  each  of  the  conductors  of  Problem  5  has  a  current  of 
50  amperes  in  the  same  direction,  what  is  the  field  at  a  point 
equidistant  from  them  ? 


32  ELECTRICAL  PROBLEMS 

8.  In  Problem  7  what  is  the  field  intensity  at  the  equidis- 
tant point  if  2  of  the  conductors  have  currents  of  50  amperes 
in  the  same  direction  and  the  third  the  return  current  of  100 
amperes  ? 

It  can  be  shown  that  the  magnetic  field  at  the  center  of  a 
very  short  circular  helix  of  n  convolutions  and  a  radius  of  r 
centimeters  is  .2  TTH  times  the  current  in  amperes  divided  by 
the  radius. 


9.  What  is  the  magnetic  force  at  the  center  of  a  circular 
convolution  with  a  radius  of  2  centimeters,  carrying  a  current 
of  4  amperes  ? 

10.  A  second  convolution  in  the  same  plane  as  and  concentric 
with  that  of  Problem  9  has  a  diameter  of  10  centimeters.     How 
large  a  current  must  flow  in  this  outer  conductor  to  double  the 
field  at  the  center? 

11.  A  third  conductor  in  the  same  plane  as  and  concentric 
with  those  of  Problem  10  has  a  current  flowing  in  the  opposite 
direction  to  those  of  the  other  convolutions,  and  exactly  neu- 
tralizing their  effect.     If  the  current  in  the  third  convolution 
is  6  amperes,  what  is  its  radius  ? 

12.  What  is  the  field  at  the  center  of  the  coil  of  a  tangent 
galvanometer  of    25    convolutions   and    20    centimeters    mean 
radius  when  traversed  by  a  current  of  .25  ampere? 

13.  What  will  be  the  deflection  of  the  needle  of  the  above 
galvanometer  when  placed  with  the  plane  of  its   coil  in  the 
magnetic  meridian,  at  a  place  where  the  horizontal   intensity 
of  the  earth's  field  is  .175  gauss? 

NOTE.  —  A  field  of  1  gauss  exerts  a  force  of  1  dyne  on  a  unit  pole.  A 
field  of  strength  H  exerts  a  force  of  Hm  dynes  on  a  pole  of  strength  m. 

SUGGESTION.  —  In  Fig.  1  let  CC\  represent  the  coil,  He  the  horizontal 
component  of  the  earth's  field,  Hg  the  magnetic  field  due  to  the  current, 
NS  the  needle,  I  its  length,  m  the  strength  of  its  pole,  and  0  the  angle  of 
its  deflection. 


THE   MAGNETIC  FIELD  DUE  TO  A  CUKKENT       33 


The  moment  of  the  couple  due  to  the  action  of  the  current  on  the 

magnet  is  Hglm  cos  6  — •  Im  cos  0  when  /  is  expressed  in  amperes. 

The  moment  of  the  couple  due  to  the  earth's  field  upon  the  magnet  is 
Hclm  sin  0. 

Since  these  are  the  only  forces  acting  on  the  needle,  they  must  balance 
if  the  deflection  is  to  be  constant  ;  hence  -  —  Im  cos  0  =  Helm  sin  0, 

and  /  =  — He  tan  0,  or  tan  0  =  '"  „.     •         —   is  called  the  constant  of 

.2  TTH  rHe          .2  irn 

the  galvanometer. 

14.  What  is  the  horizontal  intensity  of  the  earth's  field  when, 
under  the  conditions  of  Problem  12,  the  needle  is  deflected  55°? 

15.  A  tangent  galvanometer  of  100  con- 
volutions, the  mean  diameter  of  whose  coil 
is  1  meter,  has  the  plane  of  its  coil  placed 
in  the  magnetic  meridian  at  a  place  where 
the  strength  of  the  earth's  field  is  .18  gauss. 
What  current  will  produce    a   deflection 
of  45°? 

16.  The  galvanometer  of  Problem  15  is 
arranged  so  that  the  whole  instrument  may 
be  rotated  about  a  vertical  axis  till  the  zero 
of  its  scale  is  again  brought  under  the 

needle.  With  what  current  will  it  be  necessary  to  rotate  the 
instrument  30°?  What  will  a  galvanometer  designed  to  be  so 
used  naturally  be  called  ? 

17.  What  is  the  greatest  current  that  the  above  instrument, 
used  as  in  Problem  16,  will  measure? 

It  is  shown  in  mathematical  treatises  on  electricity  that  the 

magnetic  field  within  and  near  the  center  of  a  long  straight 

solenoid  of  a  uniform  number  of  turns  per  unit  length  is  4  TT 

times  the  current  multiplied  by  the  turns  per  unit  of  length. 

AirNI 


FIG.  1 


I 


where  N  is  the  whole  number  of  convolutions  of 


the  solenoid,  I  its  length  in  centimeters,  and  /  the  current  in 
amperes. 


34  ELECTRICAL  PROBLEMS 

18.  Find  the  magnetic  field  near  the  center  of  a  long  helix  of 
4  convolutions  per  centimeter  of  length  when  traversed  by  a 
current  of  15  amperes. 

19.  A  solenoid  100  centimeters  long  and  wound  with  three 
layers  of  wire  of  540  convolutions  each  is  traversed  by  a  cur- 
rent of  6  amperes.     What  is  the  magnetic  field  near  the  center 
of  the  helix  ? 

20.  What  will  be  the  flux  density  B  at  the  center  of  a  very 
long  iron  rod,  whose  permeability  is  150,  placed  in  the  axis  of 
the  helix  of  Problem  19,  the  current  being  as  before,  6  amperes? 

21.  What  is  the  flux  density  within  the  wooden  core  of  a 
helix  of  513  convolutions  per  30  centimeters  of  length,  with 
a  current  of  5  amperes  ? 

22.  Find  the  magnetic  force  within  a  helix  of  100  convolu- 
tions per  foot  of  length,  with  a  current  of  20  amperes. 

23.  A  circular  solenoid  is  formed  by  winding  wire  upon  a 
wooden  ring  whose  diameter  of  section  is  2  centimeters,  the 
inside  diameter  of  the  ring  being  10  centimeters  and  275  convo- 
lutions being  required  to  cover  the  ring.  What  is  the  magnetizing 
force  within  the  wooden  core  per  ampere  of  current  ? 

24.  What  is  the  total  magnetic  flux  per  ampere  within  the 
solenoid  ? 

25.  If  for  the  wooden  core  be  substituted  a  wrought  iron 
core  with  a  permeability  of  1500,  with  a  current  of  1  ampere, 
what  will  be  the  flux  density  ?  what  the  total  flux  ? 

26.  What  is  the  mean  length  of  flux  path  in  Problem  25  ? 
Magnetic  reluctance  is  the  analogue  of  electrical  resistance. 

The  reluctance  of  the  whole   or  of  any  part  of  a  magnetic 
circuit  varies  directly  as  the  length  and  inversely  as  the  area 

of  the  section. 

reluctivity  X  length 


reluctance  = 


area 


In  place  of  reluctivity  its  reciprocal,  permeability,  is  gener- 
ally used,  so  that  reluctance  = The  total  reluctance  of  a 


THE  MAGNETIC  FIELD  DUE  TO  A  CURRENT       35 

magnetic  circuit  formed  of  parts  arranged  in  series  is  the  sum 
of  the  reluctances  of  the  component  parts. 

We  have  seen  that  the  magnetizing  force  near  the  center  of 
a  long  helix,  or  anywhere  along  the  axis  of  a  uniformly  wound 

N 
circular  solenoid,  is  ATT  —  I.     In  the  core  of  a  circular  solenoid, 

6 

or  in  a  coil,  the  product  of  this  force  into  the  length  of  the  coil 
is  called  the  line  integral  of  magnetic  force,  or,  more  often,  the 
magneto-motive  force  of  the  coil.  Hence  the  magneto-motive 
force  of  a  magnetizing  coil  is  equal  to  .4  TT  times  the  ampere 
turns  of  the  coil,  F,  in  gilberts,  equals  HI. 

The  total  flux  is  the  ratio  of  the  magneto-motive  force  to 

magnetic  reluctance.     Flux,  <3>,  in  maxwells,  equals  -  -  •  •>  and 

(/ 

pA 

when  the  magnetic  circuit  consists  of  parts,  as  in  a  dynamo, 
we  have 


27.  How  long  an  air  gap  must  be  cut  in  the  iron  ring  of 
Problem  25  to  double  the  reluctance  of  the  whole  circuit? 

28.  If  a  certain  part  of  the  ring  of  Problem  25  consists  of  an 
alloy  of  iron  having  a  permeability  of  10,  how  long  must  the 
section  of  inferior  permeability  be  to  double  the  reluctance  of 
the  whole  circuit  ? 

29.  With  a  magneto-motive  force  of  100  gilberts,  what  is  the 
total  flux  along  an  air  path  2  centimeters  square  and  50  centi- 
meters long  ? 

30.  An  anchor  ring  whose  circular  axis  is  50  centimeters  long 
is  wound  with  200  convolutions  of  wire  carrying  a  current  of 
2  amperes.     What  is  the  magneto-motive  force  along  the  axis  ? 

31.  A  circular  solenoid  has  a  core  formed  of  a  nickel  ring. 
What  will  be  the  flux  density  per  ampere  turn  per  unit  length, 
assuming  a  permeability  of  100? 


36  ELECTRICAL  PROBLEMS 

32.  A  complete  magnetic  circuit  is  divided  into  5  sections,  as 
follows:  Section  A,  soft  iron  8  centimeters  long  and  10  x  30 
centimeters  in  section ;  Section  B,  soft  iron  21  centimeters  long 
and  15  x  35  centimeters  in  section ;  Section  C,  cast  steel  30 
centimeters  long  and  with  a  circular  section  of  25  centimeters 
diameter ;     Section    D,    cast    iron    50    centimeters    long    and 
25  X  40  centimeters  in  section ;  Section  E,  air  .75  centimeter 
long  and  18  X  40  centimeters  in  section.     What  are  the  flux 
densities  in  the  various  sections,  the  total  flux  being  6,000,000 
maxwells  ? 

33.  Under  the  conditions  of  Problem  32,  the  permeabilities 
of  such  grades  of  iron  and  steel  as  are  now  used  in  the  con- 
struction   of   electrical    machinery  are    likely   to   be   about    as 
follows:  in  ^, /A  =  63;  in  .#,/,t  =  1760;  in  <7,/z  =  925;  inD,yu=157. 
How  many  gilberts  will  be  required  for  each  section? 

34.  Around  Section  C  is  wound  a  coil  of  3750  convolutions. 
What  current  must  flow  in  this  coil  to  furnish  the  requisite 
magneto-motive  force  for  the  preceding  problems  ? 

35.  With  conditions  as  in  Problem  34,  what  current  would 
be  required  if  the  air  gap  were  eliminated  ? 

Faraday  experimentally  demonstrated  that  the  whole  quantity 
of  electricity  set  in  motion  by  the  insertion  or  removal  of  <£> 
lines  of  magnetic  induction  in  or  from  a  coil  of  N  convolutions 
forming  part  of  a  circuit  of  R  units  resistance  is  expressed  by 

$>N 

the  equation  Q  — Clearly  reversing  the  flux  in  a  certain 

-Ll? 

coil  is  equivalent  to  the  insertion  or  removal  of  twice  that  flux. 
3>N  is  called  the  total  number  of  linkages  of  current  and  mag- 
netic flux.  The  unit  of  quantity  is  the  amount  of  electricity 
conveyed  by  unit  current  in  unit  time.  The  practical  unit  of 
quantity  is  the  ampere  per  second  or  coulomb. 

If  a  secondary  coil  of  N2  convolutions  is  placed  inside  a  long 
helix  of  JVj  convolutions  per  unit  length,  the  total  change  of 
induction  in  the  secondary  at  each  make  or  break  of  the  pri- 
mary current  is  expressed  by  the  product  of  the  flux  density 


THE  MAGNETIC   FIELD  DUE  TO  A  CURRENT      37 

inside  the  long  helix  and  the  integrated  area  of  all  the  sec- 
ondary convolutions. 


36.  In  Problems  18,  21,  and  22   what  are  the  total  changes 
of  induction  in  a  secondary  of  100  convolutions,  each  3  centi- 
meters in  diameter,  on  breaking  the  primary  current  ? 

37.  In  each  case  of  Problem  36  what  quantity  of  electricity 
will  be  set  in   motion   in  a  secondary  circuit  of  1000  ohms 
resistance  by  reversing  the  primary  current  ? 

SUGGESTION.  —  Since  an  ohm  is  10°  absolute  units  of  resistance  and 
an  ampere  is  lO"1  absolute  units  of  current,  when  11  is  taken  in  ohms 
and  /  in  amperes,  3>N2  must  be  multiplied  by  10  ~8  to  give  Q  in  coulombs. 

38.  On  reversing  the  primary  current  in  Problem  21,  what 
quantity  of  electricity  is  set  in  motion  in  a  secondary  of  50 
convolutions,  2  centimeters  in  diameter,  with  a  total  resistance 
of  5000  ohms  in  the  secondary  circuit? 

39.  An  iron  ring  having  a  mean  diameter  of  20  centimeters 
and  a  cross  section  of  10  square  centimeters  has  100  turns  of 
primary  winding    and    10    turns    of   secondary.     Assuming   a 
permeability  of  1250  for  the  iron,  what  quantity  of  electricity 
is  set  in  motion  in  the  secondary  circuit  of  1500  ohms  resist- 
ance in  reversing  a  current  of  5  amperes  in  the  primary  ? 

40.  A  ballistic    galvanometer   in    the   secondary  circuit  of 
Problem  38  gives  a  deflection  of  150  scale  divisions.     What 
deflection  will    the    same    galvanometer  give   in   the    case  of 
Problem  39?     In  each  case  the  resistance  of  the  galvanometer 
is  included  in  the  resistance  given  and  the  deflection  of  the 
galvanometer  is  assumed  to  be  proportional  to  the  quantity  of 
electricity  discharged  through  it, 


CHAPTER   IX 
INDUCTANCE 

The  inductance  or  coefficient  of  self-induction  of  a  coil  is  the 
integral,  for  all  the  convolutions,  of  the  magnetic  flux  set  up 
through  each  convolution  per  unit  current  in  the  coil. 

Thus  the  inductance  of  a  coil  of  1000  convolutions  through 
each  of  which  1,000,000  maxwells,  or  lines  of  magnetic  flux, 
are  set  up  by  unit  current  or  10  amperes  is,  in  absolute  measure, 
1,000,000,000,  or,  in  the  practical  system  of  units,  1  henry. 
The  henry  is  accordingly  the  inductance  of  a  coil  in  which 
100,000,000  linkages  of  current  and  magnetic  flux  are  pro- 
duced by  each  ampere  of  current. 

It  is  frequently  difficult  and  sometimes  impossible  to  calcu- 
late the  inductance  of  coils  without  iron  cores,  but  with  com- 
plete iron  magnetic  circuits  so  arranged  that  it  may  be  assumed 
that  all  the  flux  lines  pierce  all  the  convolutions,  the  calcu- 
lation becomes  comparatively  simple,  if  the  permeability  of  the 
iron  is  known.  Thus  with  a  ring  of  soft  iron  of  mean  circum- 
ference £,  area  of  Section  A  and  permeability  /&,  overwound 
with  a  layer  of  copper  wire  of  N  turns  through  which  there  is  a 

.4  irNIfiA 
current  of  I  amperes,  we  have  the  magnetic  flux  <P  =       — j — 

L  is  used  as  the  symbol  of  inductance,  and  by  definition 

N$>  x  10-s 
L,  in  henrys,  =          — — 

Air^tiA  Xl0~'s 
hence  L  =  — —         —  • 

L 

The  quotient  of  the  inductance  of  a  coil  or  circuit  by  its 
resistance  is  called  its  time  constant,  T.  Physically  the  time 

38 


INDUCTANCE  39 

constant  represents  the   time  in  seconds   from   the   instant  of 

e  —  1 

closing  the  circuit  required  for  the  current  to  attain  the  - 

v  ^ 

part  of  its  final  value  — »  where  e  =  2.718  is  the  base  of  the 

A 

Napierian  system  of  logarithms.     Thus  in  a  circuit  of  2  henrys 
inductance  and  1  ohm  resistance  T  =  — ,   or  2 ;    that  is,  2  sec- 

R  -J    71  Q 

onds  after  closing  the  circuit  the  current  will  be  0  _1     =  .632 

^.7  lo 
of  its  final  value. 

It  is  shown    in    mathematical    treatises    on   electricity  that 

7?  t 

i  =  —  (1  —  e~r),  where  i  is  the  value  of  the  current  t  seconds 
R 

after  closing  the  circuit. 

PROBLEMS 

1.  Find  the  inductance  of  a  primary  coil  of  100  convolutions 
and  secondary  of  10   convolutions,   wound  on  an  iron  ring  of 
20  centimeters  mean  diameter  and  10  square  centimeters  sec- 
tion, the  permeability  of  the  iron  being  700. 

2.  A  transformer  has  an  iron  core  of  292  square  centimeters 
area  and  47  centimeters  in  length.      There  are  220  turns  in  the 
primary  and  22  turns  in   the   secondary.     Assuming  the  per- 
meability of  the  iron  to  be  2000,  what  are  the  inductances  of 
the  two  circuits  ? 

3.  If  in  Problem  2  the  magnetic  density  had  been  raised  to 
such   a  point   as   to   halve  the  permeability  of  the  iron,  what 
would  then  have  been  the  values  of  the  primary  and  secondary 
inductances  ? 

SUGGESTION.  —  It  should  be  noted  that  while  the  inductance  is  constant 
in  coils  without  iron,  it  is  a  variable  quantity  depending  on  the  per- 
meability, and  therefore  on  the  flux  density,  in  coils  with  iron  cores. 

4.  An  iron  ring  has  a  mean  diameter  of  15  centimeters  and 
a   circular   section   with  a  radius  of  1  centimeter.     On  it  are 
wound  2  coils  with  350  and  1000  convolutions,  respectively. 


40  ELECTRICAL  PROBLEMS 

If  the  permeability  of  the  core  is  650,  what  are  the  inductances 
of  the  two  coils  ? 

5.  A  transformer  is  wound  with  50  turns  in  the  primary  coil 
and  1000  in  the  secondary.     The  magnetic  circuit  has  a  mean 
length  of  50  inches  and  an  area  of  12  square  inches.     Assuming 
a  permeability  of  1800,  what  are  the  inductances  of  the  two 
coils  ? 

6.  An  iron  ring  has  a  mean  diameter  of  25  centimeters  and 
a  circular  section  of  1.5  centimeters  radius.      On  it  are  wound 
2  coils,  a  primary  of  425  convolutions  and  a  secondary  of  17 
convolutions.     Find  the  inductance  of  each  of  the  coils  when 
the  permeability  of  the  iron  core  is  775. 

7.  A  certain  transformer  has  a  magnetic  circuit  whose  mean 
length  is  35  inches,  and  whose  gross  section  is  40  square  inches, 
10%   of  which  is  insulation.     The  primary  coil  contains  175 
convolutions  and  the  secondary  35  convolutions.     If  the  per- 
meability of  the  iron  is  1250,  what  are  the  inductances  of  the 
two  coils  ? 

8.  A  reactance  coil  is  formed  of  a  ring  of  soft  iron  wire  17.5 
square  centimeters  in  section,  and  with  a  mean  circumference 
of  50  centimeters,  overwound  with  650  convolutions  of  copper 
wire.     Assuming  the  iron  to  have  a  permeability  of  900,  what 
is  the  self-induction  of  the  coil  ? 

9.  Plot  a  curve  showing  the  growth  of  current  for  7  seconds 
after  closing  a  circuit  of  1  henry  inductance  and  1  ohm  resist- 
ance, upon  which  is  impressed  an  electro-motive  force  of  1  volt. 

|  =  i,,-=|(i_,-f)=i-i.      . 

When   *  =  .!,  -=.9048,  i  =  l  (1-.9048)  =  .0952,  when  t  =  .2,  i  =  .1813,  etc. 

10.  Plot  a  curve  showing  the  rise  of  current  in  a  circuit  of 
4  henrys  inductance   and  3   ohms  resistance,  upon   which  is 
impressed  an  electro-motive  force  of  10  volts. 

11.  If  1000  volts  are  impressed  on  a  circuit  of  10  henrys 
inductance  and  4  ohms  resistance,  what  will  be  the  value  of  the 


INDUCTANCE 


41 


current  at  the  end  of  T^-7  second  ?  what  at  the  end  of  1  second  ? 
what  at  the  end  of  5  seconds  ?  what  the  final  value  ? 

12.  A    circuit   contains    .2    henry   inductance    and   5    ohms 
resistance.     What  is  the  value  of  the  time  constant  and  what 
will  be  the  value  of  the  current  T  seconds  after  applying  an 
electro-motive  force  of  60  volts  ? 

13.  A  circuit  contains  10  ohms  resistance  and  has  impressed 
on  it  an  electro-motive  force  of  100  volts.     Draw  curves  of 
current  at  the  end  of  .5,  1,  and  2  seconds  as  the  inductance 
varies  from  0  to  20  henry s. 

14.  A  circuit  of  50  ohms  resistance  has  impressed  upon  it  an 
electro-motive  force  of  100  volts.    Plot  curves  showing  the  rise 
of  current  when  the  circuit  contains  inductances  of  25  henrys, 
50  henrys,  and  100  henrys,  respectively. 

15.  A  circuit  with  50  henrys  inductance  has  impressed  upon 
it  an  electro-motive  force  of  100  volts.     Plot  curves  showing 
the  rise  of  current,  with  resistances  of  25  ohms,  50  ohms,  and 
100  ohms,  respectively. 

16.  The  instantaneous  values,  i,  of  the  current  through  the 
field  magnet  coils  of  a  dynamo  were  noted  at  various  intervals 
of   time,  tj  after   closing   the    field   magnet   circuit,  with   the 
following  results : 

t  in  seconds        i  in  amperes 


1.0 
1.5 
2.0 
2.5 
3.0 
3.5 


2.1 

3.3 

3.9 

4.3 

4.7 

5.05 

5.3 


t  in  seconds 

i  in  amperes 

4 

5.45 

5 

5.63 

6 

5.73 

7 

5.81 

8 

5.83 

60 

5.84 

Plot  the  curve  showing  the  rise  of  current  in  the  field  coils, 
and  from  it,  by  observing  T,  find  the  inductance,  the  resistance 
being  13.2  ohms. 

If  we  neglect  the  falling  off  in  the  flux  density  near  the  ends, 
we  may  calculate  the  inductance  of  a  long  helix  without  an 


42  ELECTRICAL  PROBLEMS 

iron  core  by  the  same  formula  used  for  a  coil  with  a  complete 
iron  circuit.     The  field  inside  the  long  helix  is  expressed  by 


.4  irNI       ,    ,  A  irNAI    ,  . 

H—  -  and  4>  —HA  =  -  ;  hence  L  =  ---  x  10  ~\ 

L  (/  L 

where  r  is  the  radius  of  the  helix. 

For  a  circular  coil  of  radius  very  large  compared  with  its 
section,  such,  for  instance,  as  the  coil  of  a  tangent  galva- 
nometer, the  inductance  is  approximately  expressed  by 

L  =  4  irN*r  (log,  —  -  2 

where  r  is  the  radius  of  the  coil  and  a  the  side  of  the  approxi- 
mately square  section. 

When  2  coils  are  so  situated  that  a  current  of  1  ampere  in 
either  sets  up  100,000,000  linkages  of  magnetic  flux  and  con- 
volutions in  the  other,  they  are  said  to  have  a  mutual  induc- 
tance, Jf,  of  1  henry.  In  the  absence  of  iron  mutual  inductance 
as  well  as  self  inductance  is  constant. 

The  mutual  inductance  of  2  concentric  helices  of  nearly  the 
same  radius  will  be  approximately 

_  47rV7VyV2xlQ--9 

where  r  is  the  radius  of  the  inner  coil  and  Nr  and  N.2  are  the 
numbers  of  convolutions  in  the  2  coils,  respectively. 

The  value  of  M  for  2  short  coaxial  coils  situated  in  the  same 
plane  and  of  nearly  equal  radii,  r  and  r  +  c,  respectively,  is 
approximately  expressed  by 

ge  —  - 

or  if  the  plane  of  one  of  the  coils  is  moved  in  the  direction  of 
its  axis,  so  that  the  constant  distance  between  the  mean  circular 
axes  is  6,  then 

M=  4  TTN^T  Aog,       -  2    10-9. 


INDUCTANCE  43 

17.  What  is  the  inductance  of  a  coil  of   100  convolutions 
wound   on   a   glass  rod   of  2    centimeters  radius,  the  winding 
occupying  a  length  of  25  centimeters  ? 

18.  A  solenoid   with    200   convolutions  in  a  length   of  30 
centimeters  is  wound  on  a  wooden  cylinder  having  a  radius  of 
3  centimeters.     What  is  its  inductance  ? 

19.  What  is  the  inductance  of  a  solenoid  of  5000  convolu- 
tions in  a  length  of  13  feet,  wound  on  a  cylinder  of  wood  1 
foot  in  diameter? 

20.  What  is  the  mutual  inductance  of  two  concentric  coils 
65  centimeters  long,  of  200  and  630  convolutions,  respectively, 
the  inner  coil  being  wound  directly  on  a  paper  tube  16  centi- 
meters in  diameter  ? 

21.  What  is  the  inductance  of  the  coil  of  a  tangent  galva- 
nometer consisting  of  100  convolutions  of  a  mean  radius  of  1 
meter,  wound  in  a  groove  1.5  centimeters  square? 

22.  Two  circular  coils  are  wound  in  parallel  grooves  1  centi- 
meter apart.    Each  coil  has  a  mean  radius  of  25.75  centimeters, 
a  section  1.3  centimeters  square,  and  contains  234  convolutions. 
What  is  their  mutual  inductance  ? 

23.  What  is  the  inductance  of  the  2  coils  of  Problem  22 
when  joined  in  series  ? 

SUGGESTION.  —  The  inductance  of  2  coils  in  series  is  the  sum  of  their 
separate  inductances  plus  twice  their  mutual  inductance. 

24.  A  groove  with  a  rectangular  section  1  centimeter  wide 
and   2   centimeters  deep  is  turned  in  a  wooden  cylinder,  the 
radius  of  the  bottom  of  the  groove  being  20  centimeters.     In 
this  groove  is  wound  a  primary  coil  of  25  convolutions,  having 
a  depth  of  1   centimeter  and  a  secondary  coil   of  400   convo- 
lutions also  having  a   depth  of   1   centimeter.     What  is  their 
mutual  inductance?     What  is  the  inductance  of  the  2  coils 
joined  in  series? 

A  very  important  calculation  is  that  of  the  inductance  of 
transmission  lines.     The  formula  for  the  inductance,  per  unit 


44  ELECTRICAL  PROBLEMS 

length,  of  straight  parallel  conductors  carrying  equal  but  oppo- 
sitely directed  currents  is 

L  = 

in  which  /A  is  the  permeability  of  the  medium,  /*'  that  of  the 
wires,  d  the  distance  between  the  axes  of  the  wires,  and  r  the 
radius  of  the  wires. 

In  case  of  non-magnetic  wires  hung  in  air,  the  inductance  in 
henrys  for  the  length  I  is  expressed  by  the  equation 

_L/  == 


25.  What  is  the  inductance  per  kilometer  of  a  line  of  2  copper 
wires  45  centimeters  apart  and  1  centimeter  in  diameter? 

26.  What  is  the  inductance  of  a  line  200  miles  long  consist- 
ing of  2  wires  15  inches  apart  and  .1616  inch  in  diameter? 

27.  What  is  the  inductance  per  mile  of  2  No.  4  B.S.G.  copper 
wires  strung  18  inches  apart? 


CHAPTER    X 
THE  CONDENSER 

The  capacity  of  a  condenser  varies  inversely  with  the  thick- 
ness of  the  dielectric  and  directly  with  the  area  of  the  conduct- 
ing surfaces  and  the  permittivity,  or  dielectric  constant,  formerly 
known  as  the  specific  inductive  capacity,  of  the  insulating 
material.  Expressed  in  C.G.S.  electrostatic  units, 

C  =  K-T^"> 

4  Trd 

C  being  the  capacity  in  absolute  units,  K  the  permittivity,  d  the 
thickness  of  the  dielectric,  and  A  the  effective  area  of  the  con- 
ducting plates,  usually  planes,  cylinders,  or  spheres.  The  quan- 
tity of  electricity  required  to  produce  a  difference  of  potential 
E  between  the  plates  of  a  condenser  of  capacity  C  is  Q  =  EC  ; 

whence  C  =  -  and  E  =  —. 
E  C 

The  commonly  used  unit  of  capacity  is  the  microfarad,  or 
millionth  part  of  a  farad,  the  practical  unit  of  a  farad  being  the 
capacity  that  will  be  charged  to  a  potential  of  1  volt  by  a 
quantity  of  1  coulomb,  or  by  a  current  of  1  ampere  flowing  for 
a  second  into  the  condenser. 

The  above  relations  between  Q,  E,  and  C  are  true  for  both 
the  absolute  and  the  practical  units.  The  equation 

C-K     A 
K 


becomes,  when  C  is  the  capacity  in  microfarads   and  A  and  d 
are  expressed  in  square  and  linear  centimeters,  respectively, 

A 


C=K 


4  Trd  x  9  x  106 

45 


46  ELECTRICAL  PROBLEMS 

The  following  table  gives  approximate  values  of  the  permit- 
tivity of  various  dielectrics  in  use  in  condensers : 

Air  or  vacuum 1 

Paraffined  paper 2 

India  rubber,  pure 2.34 

India  rubber,  vulcanized 2.94 

Gutta-percha 4.2 

Mica 5 

Glass  .  6 


PROBLEMS 

1.  What  charge  will  be  imparted  to  a  condenser  having  a 
capacity  of  100  units  by  a  potential  difference  of  1000  units? 

2.  How  many  volts  are  required  to  impart  a  charge  of  .001 
coulomb  to  a  capacity  of  4  microfarads  ? 

3.  What  capacity  will  hold  a  charge  of  900  units  under  a 
difference  of  potential  of  45? 

4.  What  quantity  will  charge  a  capacity  of  .01  microfarad  to 
a  potential  of  6  volts  ? 

5.  How  long  must  a  current  of  2.5  amperes  flow  to  charge 
a  capacity  of  .21  microfarad  to  a  potential  of  .6  volt  ? 

6.  What  is  the  capacity  of  a  condenser,  formed  of  2  plates 
1  meter  square,  separated  by  1  millimeter  of  air? 

7.  Of  how  many  plates  1  meter   square,  separated  by  1.3 
millimeters  of  air,  and  with  the  2  external  plates  of  the  same 
polarity,  must  a  condenser  be  composed  to  have  a  capacity  not 
less  than  .4  microfarad? 

8.  What  must  be  the  area  of  the  plates  in  Problem  7  if  the 
condenser  is  to  have  a  capacity  of  precisely  .4  microfarad  ? 

9.  If  the  condenser  of  Problem  8  have  its  plates  separated 
by  mica  .5  millimeter  thick,  what  is  its  capacity  ? 

10.  What  is  the  capacity  of  a  condenser  built  of  200  sheets 
of  tin  foil,  each  100  square  centimeters  in  effective  area,  sepa- 
rated by  paraffined  paper  .1  millimeter  thick? 


THE  COKDENSEK  47 

11.  A  condenser  is  to  be  built  of  sheets  of  tin  foil  having  an 
effective  area  8  centimeters  square,  separated  by  .5  millimeter 
of  pure  rubber.     What  number  of  plates  will  give  the  nearest 
approach  to  .01  microfarad? 

12.  What  will  be  the  exact  capacity  of  the   condenser  of 
Problem  11? 

13.  Keeping  the  tin  foil  sheets  of  the  condenser  of  the  previ- 
ous problem  8  centimeters  wide,  what  must  be  their  effective 
length  to  produce  a  capacity  exactly  .01  microfarad? 

14.  A  condenser  is  built  of  160  circular  sheets  of  tin  foil 
separated  by  mica  .5  millimeter  thick.     How  many  tin  foil  sur- 
faces will  be  of  one  polarity  ?    What  must  be  the  diameter  of  the 
sheets  that  the  condenser  may  have  a  capacity  of  i  microfarad  ? 

Two  concentric,  cylindrical,  metallic  surfaces  insulated  from 
each  other  may  serve  as  a  condenser.  The  equation  for  capacity 
becomes 

C  =  K 


in  absolute  electrostatic  units,  where  I  is  the  length  of  the 
metallic  cylinders,  R  the  internal  radius  of  the  outer,  and  r 
the  external  radius  of  the  inner.  In  microfarads 


10'kfol 

?,  R,  and  r  being  expressed  in  centimeters  as  in  the  formula  for 
the  capacity  in  absolute  units. 

15.  What  is  the  capacity  in  microfarads  per  mile  of  a  copper 
wire  .4  inch  in  diameter,  contained  within  and  separated  from 
a  hollow  metallic  cylinder  by  .06  inch  of  vulcanized  rubber? 

16.  The  outer  tube  of  Problem  15  is  .1  inch  thick  and  is 
covered  with  .06  inch  of  gutta-percha.     If  the  whole  is  immersed 
in  water,  what  is  the  capacity,  in  microfarads  per  mile,  of  the 
condenser  formed  by  the  outer  tube  and  the  water? 


48  ELECTRICAL  PKOBLEMS 

17.  A  wire  4  millimeters  in  diameter  and  insulated  with 
5  millimeters  of  gutta-percha  is  placed  in  water.  What  is  the 
capacity  in  microfarads  per  kilometer  ? 

It  is  shown  in  mathematical  treatises  on  electricity  that  the 
charge  q  in  a  condenser  of  capacity  Cy,  t  seconds  after  being 
connected  with  a  source  of  constant  potential  JK,  is  expressed  by 

q  =  EC  (1  -  «*'<>),  or  q  =  Q  fl  -  J 

\        eRC 

where  Q  is  the  final  charge,  K  the  resistance  in  circuit  with  the 
condenser,  and  e  =  2.718  is  the  base  of  the  Napierian  system  of 
logarithms.  R  C  is  called  the  time  constant  of  the  circuit  and 
is  the  time  in  seconds  from  the  instant  of  closing  the  circuit  to 

e  —  \ 
the  instant  when  the  charge  reaches  the  -     —  =  .632  part  of  its 

& 

final  value.  This  expression  is  true  expressed  in  either  absolute 
or  practical  units. 

1*8.  Draw  a  curve  showing  the  amount  of  charge  at  each  instant 
of  time  in  a  condenser  of  1  microfarad  capacity,  charged  through 
a  resistance  of  400  ohms  by  an  electro-motive  force  of  100  volts. 

Solution.     R  -  400,  C  =  10  ~fi,  RC  =  .0004  second. 

Q  =  EC  =  100  x  10  ~G  =  .0001  coulomb,  q  =  .0001  /I-  — —  V 

\  g.0004/ 

When  t  =  .0001,  —  =  .25  ;  q  =  .0001  (1  -  .7788)  =  .00002212. 

RC 

When  t  =  .0002,  q  =  .00003934,  etc. 

It  is  further  shown  that  where  i  denotes  the  instantaneous 

Q  L 

value  of  the  current  flowing  into  a  condenser,  i  =  — —  times  e  RC. 

_/t  L/ 

19.  Draw  the  curve  showing  the  instantaneous  values  of  the 
charging  current  in  Problem  18. 

20.  In  what  multiple  of  EC  will  a  condenser  receive  its  full 
charge  to  within  one  part  in  1000? 

Solution.  q  =  Q  -  Qe  ~  ™,  -^^   -  e  '  ™  =  — , 

t  =  RC  loge  1000  =  6.908  RC  seconds. 


THE  CONDENSER  49 

21.  In  what  multiple  of  RC  will  a  condenser  receive  its  full 
charge  to  within  one  part  in  1,000,000? 

22.  A  condenser  of  4  microfarads  capacity  is  charged  through 
a  resistance  of  1000  ohms  by  an  electro-motive  force  of  500 
volts.     Calculate  the  instantaneous  value  of  the  charge  at  the 
end  of  .001  second,  .006  second,  and  .02  second. 

23.  Calculate  the  instantaneous  values  of  the  current  flowing 
into  the  condenser,  under  the  conditions  of  and  at  the  instants 
specified  in  Problem  22. 

24.  A  condenser  of  .5  microfarad  capacity  is  charged  through 
a  resistance  of  500  ohms.    What  is  the  time  constant  Tl    With 
75  volts  impressed  upon  the  circuit,  what  will  be  the  charge  and 
what  the  current  T  seconds  after  closing  the  circuit? 

25.  A  condenser  of  2  microfarads  capacity  is  being  charged 
by  an  electro-motive  force  of  250  volts.     Plot  curves  showing 
charge  at  the  end  of  .0001  second,  .0005  second,  and  ,002  sec- 
ond as  the  resistance  varies  from  0  to  500  ohms. 

26.  Plot  curves  of  current  under  the  conditions  of  and  at  the 
instants  specified  in  Problem  25. 

27.  A  circuit  containing  a  resistance  of  200  ohms  has  im- 
pressed upon  it  an  electro-motive  force   of   100   volts.      Plot 
curves  showing  the  instantaneous  values  of  the  charge  as  the 
capacity  varies  from  0  to  5  microfarads,  at  the  end  of  .0001 
second,  .0005  second,  and  .002  second. 

28.  Plot  curves  showing  the  values  of  the  current  under  the 
conditions  of  and  at  the  instants  specified  in  Problem  27. 

The  capacity  of  an  aerial  line  I  centimeters  long,  consisting 
of  2  wires  separated  by  a  space  of  d  centimeters  and  each  r 

centimeters  in  radius,  is,  in  microfarads,  C  =  -  . 


29.  What  is  the  capacity  per  kilometer  of  a  line  of  2  copper 
wires  5  millimeters  in  diameter,  separated  by  50  centimeters  ? 

30.  What  is  the  capacity  per  mile  of  2  No.  0  B.S.G.  wires 
hung  2  feet  apart? 


50  ELECTKICAL  PKOBLEMS 

31.  What  is  the  capacity  of  100  kilometers  of  transmission 
line,  the  2  wires  being  1  meter  apart  and  each  8  millimeters  in 
diameter? 

32.  What  is  the  capacity  of  a  line  150  miles  long,  consisting 
of  2  No.  00  B.S.G.  wires  hung  5  feet  apart? 

The  joint  capacity  of  condensers  in  multiple  is  the  sum  of 
their  separate  capacities.  The  joint  capacity  of  condensers  in 
series  is  the  reciprocal  of  the  sum  of  the  reciprocals  of  their 
separate  capacities. 

33.  What  is  the  joint  capacity  of  3  condensers  of  2,  3,  and  4 
microfarads,  respectively,  all  joined  in  multiple  ? 

34.  What  is  the  joint  capacity  of  the  above  condensers  joined 
in  series  ? 

35.  Two  condensers  of  2  and  3  microfarads,  respectively,  are 
joined  in  multiple.      In  series  with  this  arrangement  is  a  con- 
denser of  4  microfarads  capacity.    What  is  the  capacity  of  the 
combination  ? 


CHAPTER   XI 
THERMOELECTRICITY 

In  a  circuit  formed  of  two  wires  of  different  metals,  one  of 
whose  junctions  is  maintained  at  a  higher  temperature  than  the 
other,  there  is  a  resultant  electro-motive  force  set  up  around  the 
circuit.  Electro-motive  forces  have  been  observed  in  nearly  all 
metals  but  lead,  when  heated  unequally  in  parts,  and  accord- 
ingly lead  has  been  selected  as  the  standard  with  which  to 
compare  other  metals  in  expressing  the  thermoelectric  power  of 
couples,  or  the  resultant  electro-motive  force  around  the  circuit 
for  a  difference  in  temperature  of  1°  C.  of  the  two  junctions  of 
the  couple.  This  thermoelectric  power  is  not  affected  by  the 
presence  of  any  conductor  used  to  join  the  ends  of  the  members 
of  the  couple,  such  as  solder  or  a  galvanometer,  and  is  dependent 
on  the  mean  temperatures  of  the  junctions. 

The  following  table  from  Gerard's  Electricity  and  Magnetism 
(page  182)  enables  one  to  calculate  the  thermoelectric  power  of 
several  metals  taken  with  lead  by  the  formula 

e  =  (  a  +  b  — - 


where  e  is  the  resultant  electro-motive  force  in  volts  around  the 
circuit,  a  and  b  physical  constants  of  the  metal  used,  and  t  and  tr 
the  temperatures  of  the  hotter  and  cooler  junctions,  respectively, 
in  degrees  Centigrade. 

Metal  a  b 

Copper  -.00000134  -.0000000094 

Iron  -.00001715  +.0000000482 

German  silver  +.00001194  +.0000000506 

51 


52  ELECTKICAL  PROBLEMS 

A  positive  value  of  e  shows  a  tendency  of  the  current  to  flow 
to  lead  across  the  warmer  junction  and  a  negative  value  of  e  a 
resultant  electro-motive  force  in  the  opposite  direction.  When 
two  metals  other  than  lead  are  used  in  the  couple  the  values  of 
a  and  b  are  the  differences  between  the  above  values  of  a  and  b 
for  the  two  metals. 

PROBLEMS 

1.  A  circuit  is  formed  of  iron  and  lead,  the  junctions  being 
kept  at  —  5°  C.  and  -f-  5°  C.,  respectively.     What  is  the  result- 
ant   electro-motive    force    around    the    circuit,    and    in    which 
direction  does  the  current  flow? 

NOTE.  —  Unless  otherwise  stated,  the  current  direction  will  be  assumed 
to  be  that  across  the  warmer  junction. 

Solution.     For  iron,  a  =  -.00001715,  b  =  +.0000000482. 
e  =  (-  .00001715  +  .0000000482  5+^~o))  (5  -  (-  5)  )  =  -.0001715  volt. 

The  current  leaves  the  lead. 

2.  If  each  of  the  junctions  of  Problem  1  has  its  temperature 
raised  10°,  what  is  the  direction  and  value   of  the   resultant 
electro-motive  force  ? 

3.  A  circuit  contains  a  piece  of  iron  and  a  piece  of  copper 
wire  soldered  together.     If  the  temperatures  of  the  junctions 
are,  respectively,  120°  C.  and  20°  C.,  what  is  the  value  and 
direction  of  the  resultant  electro-motive  force? 

4.  If  the  temperatures  of  the  junctions  in  Problem  3  are 
changed  to  420°  C.  and  0°  C.,  what  is  the  value  and  direction 
of  the  resultant  electro-motive  force  ? 

5.  A  circuit  is  formed  of  lead  and  German  silver  with  one 
junction  kept  at  0°  C.     Plot  a  curve  showing  the  electro-motive 
force  in  the  circuit  as  the  temperature  of  the  other  junction 
changes  from  0°  C.  to  250°  C. 

6.  Plot  a  curve  showing  the  variation  of  the  resultant  electro- 
motive force  in  a  copper-iron  circuit  with  one  junction  kept  in 


THERMOELECTKICITY  53 

ice  water  and  the  temperature  of  the  other  varying  from  0°  C. 
to  600°  C. 

The  thermoelectric  couple  is  frequently  used  for  the  measure- 
ment of  high  temperatures,  and  for  this  purpose  it  is  found  that 
only  platinum  and  its  alloys  of  iridium  or  rhodium  are  practi- 
cable. Le  Chatelier  and  Boudouard  give  the  following  table 
of  the  electro-motive  force  in  microvolts  set  up  in  junctions  of 
a  platinum  wire  with  one  of  platinum  alloyed  with  10%  of 
iridium  or  rhodium,  at  certain  temperatures. 

Temperature  Iridium  Rhodium 
(Degrees  C.) 

100  517  565 

448  3,228  3,450 

930  11,000  8,500 

1,500  15,100 

The  net  electro-motive  force  in  the  circuit  being  the  differ- 
ence between  those  of  the  two  junctions,  if  this  can  be  observed 
and  the  temperature  of  the  cooler  obtained,  the  temperature  of 
the  hotter  can  be  deduced  from  a  curve,  giving  the  electro- 
motive force  set  up  at  each  temperature  throughout  the  neces- 
sary range. 

This  net  electro-motive  force  can  be  obtained  in  two  ways : 
first,  by  balancing  it  against  the  drop  in  a  certain  section  of 
a  known  resistance,  R,  when  traversed  by  the  current  from  a 
standard  cell.  If  E  is  the  electro-motive  force  of  the  cell,  /  its 
current  through  the  known  resistance  R,  and  r  the  adjustable 
resistance  in  which  the  drop  balances  e,  we  have  e  =  Ir, 

n* 

E=IR,  ande  =  jEr— •     Second,  the  electro-motive  force  e  may 
R 

be  found  by  connecting  the  ends  of  the  couple  to  the  terminals 
of  a  galvanometer  whose  resistance  is  constant  and  so  large  that 
the  slightly  variable  resistance  of  the  couple,  due  to  change  in 
temperature,  may  be  neglected.  We  have,  then,  e  =  IR,  where 
R  is  the  resistance  of  the  circuit  and  I  the  current  set  up  in  it 
by  the  electro-motive  force  e. 


54  ELECTRICAL   PROBLEMS 

7.  Construct,  from  the  table  given  above,  curves  of  the  varia- 
tion with  temperature  of  the  electro-motive  force  at  a  junction 
of  a  platinum  and  a  platinum  with  10%  iridium  wire  and  at  a 
junction  of  a  platinum  and  a  platinum  with  10%  rhodium  wire. 

8.  A  junction  of  platinum  and  platinum  iridium  10%  alloy 
is  used  to  determine  the  temperature  of  a  potter's  furnace.     It 
is  connected  to  the  terminals  of  a  galvanometer  of  5000  ohms 
resistance  whose  temperature  is  18°  C.     What  current  will  flow 
if  the  temperature  of  the  furnace  is  650°  C.? 

9.  If  in  Problem  8  the  deflection  is  11.5  millimeters  of  scale, 
what  temperature  of  the  furnace  is  indicated  by  a  deflection  of 
18  millimeters? 

10.  If  the  above  couple  is  heated  to  915°  C.  and  the  termi- 
nals are  connected  to  a  coil  of  copper  of  2250  ohms  resistance 
when  kept  in  ice  water,  what  current  will  flow  ? 

11.  A  thermoelectric  couple  is  formed  of  a  wire  of  platinum 
and  one  of  platinum  and  rhodium  in  the  proportion  of  9  to  1. 
The  junction  is  heated  to  1375°  C.  and  the  terminals  in  a  room 
at  20°  C.  are  connected  to  a  galvanometer.     A  standard  cell 
giving  an  electro-motive  force  of  1.432  volts  at  this  temperature 
sets  up  a  current  of  .01  ampere  in  a  wire  of  5  ohms  resistance 
per  meter.     The  terminals  of  above  galvanometer  are  connected 
to  points  on  this  wire  so  that  the  difference  of  potential  thereby 
produced  between  the  terminals  of  the  galvanometer  opposes 
that  due  to  the  couple.     For  no  deflection,  how  far  apart  must 
the  contacts  be  ? 

12.  Later  it  is  found  necessary  to  separate  the  contacts  18 
and  then   7   centimeters.     What  temperatures    of   the    hotter 
junction  are  then  indicated? 


CHAPTER    XII 
ELECTRO-CHEMISTRY 

When  a  current  of  electricity  is  passed  through  a  chemical 
compound  in  the  liquid  state  the  liquid  is  in  general  decom- 
posed ;  the  more  electro-positive  products  of  the  decomposition, 
metals,  etc.,  being  deposited  upon  the  negative  terminal  or  elec- 
trode, and  the  more  electro-negative  products,  oxygen,  etc.,  upon 
the  positive  electrode.  This  decomposition  is  known  as  elec- 
trolysis, the  liquid  is  called  an  electrolyte,  and  the  products  of 
the  electrolysis  are  termed  ions. 

The  following  laws  were  discovered  by  Faraday : 

1.  The  amount  of  an  ion  deposited  and  of  the  electrolyte 
decomposed  is  proportional  to  the  amount  of  electricity  that  has 
traversed  the  electrolyte. 

2.  The  amount  of  electrolytic  action  is  the  same  in  each  of 
several  electrolytic  cells  arranged  in  series. 

3.  The  amount  of  an  ion  set  free  is  equal  to  the  total  quan- 
tity of  electricity  that  has  passed  through  the  electrolyte  multi- 
plied by  the  electro-chemical  equivalent  of  the  ion. 

The  following  are  the  electro-chemical  equivalents  of  some 
common  elements  in  grams  per  coulomb  of  electricity: 

Hydrogen 0.0000104 

Gold 0.00068 

Silver 0.00112 

Copper  (from  cupric  salt) 0.000328 

Nickel 0.000304 

Zinc 0.000337 

Oxygen 0.0000829 

Chlorine   .  0.000367 


56  ELECTRICAL  PROBLEMS 

PROBLEMS 

1.  A  current  of  4  amperes  flows  through  water.      At  what 
rate  are  hydrogen  and  oxygen  liberated  ? 

Solution.  Four  amperes  are  4  coulombs  a  second.  Four  coulombs 
a  second  liberate  4  x  .0000104  =  .0000416  gram  of  hydrogen  and 
4  x  .0000829  =  .0003316  gram  of  oxygen  per  second. 

2.  If  the  current  of  Problem  1  is  passed  through  hydrochloric 
acid,  what  elements  will  be  liberated,  and  at  what  rate  ? 

3.  From  a  solution  of  copper  sulphate  how  much  copper  will 
be  deposited  per  hour  by  a  current  of  1  ampere  ? 

4.  Two  electrolytic  baths  are  arranged  in  series  in  a  circuit, 
one  containing  copper  sulphate  and  the  other  silver  nitrate.     If 
40  grams  of  copper  are  deposited  hourly,  what  is  the  current, 
and  how  much  silver  is  deposited  in  10  minutes  ? 

5.  A  piece  of  metal  weighing  400  grams  is  to  be  plated  with 
5J&  of  its  weight  of  gold.     It  is  placed  in  a  solution  of  gold 
chloride,  and  a  current  of  10  amperes  is  used.    How  long  must 
it  remain  in  the  bath  ? 

6.  A  casting  is  to  have  deposited  on  it  2  pounds  copper  and 
then  2  pounds  nickel.     It  is  to  be  left  in  the  copper  bath  for 
10  hours.     With  the  same  current  how  long  must  it  remain  in 
the  nickel  bath  ? 

7.  A  current  of  20  amperes  is  divided  between  two  branches, 
one  containing  an  electrolytic  gold-plating  bath,  the  other  a  ves- 
sel of  water  to  be  decomposed.     The  current  divides  so  that 
equal  weights  of  oxygen  and  gold  are  separated.     What  are  the 
currents,  and  how  much  hydrogen  and  oxygen  are  separated  ? 

8.  In  a  plant  for  electrolytically  refining  copper  a  current  of 
1000  amperes  flows  23  hours  per  day,  300  days  per  year,  through 
60  cells   arranged  in  series.      How  many  tons  of  copper  are 
refined  each  year  ? 

9.  A  bath  for  the  electric  deposition  of  copper  runs  8  hours 
per  day,  with  a  current  of  500  amperes.     How  long  must  it  run 
to  deposit  enough  copper  for  1  mile  of  No.  00  B.S.G.  wire? 


ELECTRO-CHEMISTRY  57 

10.  A  copper  voltameter,  consisting  of  5  cells  in  multiple, 
is  connected  in  series  with  an  ammeter,  a  variable  resistance, 
and  a  storage  battery.     A  constant  current  is  maintained  for 
exactly  1  hour,  when  the  gain  in  grams  in  the  weight  of  the 
cathodes  was   found  to  be,  respectively,  1.536,  1.473,  1.621, 
1.574,    and    1.512.      What    should    have    been    the    ammeter 
reading  ? 

11.  A   constant  current  deposits   3.628  grams  of  silver  in 
45  minutes.     What  is  the  value  of  the  current? 


CHAPTER    XIII 
ALTERNATING   ELECTRO-MOTIVE   FORCES  AND   CURRENTS 

Alternating  or  periodic  electro-motive  forces  and  currents 
may  be  represented  by  clock  diagrams,  in  which  the  instanta- 
neous value  of  the  periodic  function  is  represented  by  the  pro- 
jection of  a  uniformly  rotating  vector  upon  the  diameter  of  a 
circle,  or  by  wave  diagrams  in  which  the  instantaneous  value 
of  the  function  is  represented  by  an  ordinate  progressing  with 
uniform  velocity  along  the  axis  of  abscissas,  whose  length  at 
any  instant  is  the  above-described  projection. 

A  review  of  the  methods  of  generating  electro-motive  forces 
and  currents  will  sho-w  that  the  positive  and  negative  half  waves 
must  generally  be  similar  in  shape.  Frequently  these  curves 
approximate  to  sine  curves,  and  being  periodic  they  may,  by 
Fourier's  theorem,  be  considered  as  the  resultant  of  two  or  more 
sine  waves. 

Even  where  they  differ  markedly  from  a  simple  sine  curve 
there  is  always  a  simple  sine  wave  of  the  same  periodicity, 
called  the  equivalent  sine  wave,  that  may  for  most  purposes  be 
substituted  for  the  actual  periodic  curve. 

Since  electrical  energy  varies  as  the  square  of  the  electro- 
motive force,  or  as  the  square  of  the  current,  alternating 
electro-motive  forces  and  currents  are  measured  by  the  square 
root  of  the  mean  of  the  squares  of  the  equidistant  ordinates  of 
the  particular  wave  diagrams.  This  V mean2  (square  root  of 
the  mean  square)  value  is  called  the  virtual  value  of  the  electro- 
motive force  or  current  as  the  case  may  be,  and  all  alternating 
current  ammeters  and  voltmeters  are  made  to  indicate  virtual 
values. 

58 


ELECTRO-MOTIVE  FORCES   AND  CURRENTS        59 

The  equation  for  a  harmonically  varying  electro-motive  force 
is  e  =  E  sin  cot,  where  e  is  the  instantaneous  value  of  the 
electro-motive  force,  E  the  maximum  value ;  GO,  numerically 
equal  to  2  TT  times  the  periodicity,  is  the  linear  velocity  of 
the  ordinate  of  the  sine  curve,  or  the  angular  velocity  of  the 
vector  in  the  clock  diagram,  and  t  is  the  time  in  seconds  from 
the  beginning  of  the  reckoning  to  that  instant  at  which  the 
electro-motive  force  has  the  value  e. 

The  equation  for  a  periodic  but  non-harmonic  electro-motive 
force  will  vary  from  the  above  only  in  containing  more  than 
one  term  on  the  right-hand  side.  Nearly  all  the  electro-motive 
forces  met  with  in  practice  may  be  very  approximately  expressed 
by  the  following  equation: 

e  =  El  sin  cot  +  E3  sin  (3  cot  —  03)  +  E5  sin  (5  cot  —  05), 

in  which  03  and  #5  represent  angles  by  which  the  zero  of  the  par- 
ticular harmonic  is  displaced  from  the  zero  of  the  fundamental. 
The  fact  that  the  positive  and  negative  half  waves  are  usually 
similar  in  shape  shows  that  commonly  only  the  odd  numbered 
harmonics,  3  a),  5  o>,  etc.,  occur  in  practice.  The  higher  harmonics 
E1  sin  (7  cot  —  07)  +  EQ  sin  (9  cot  —  09),  etc.,  when  they  occur,  are 
usually  too  small  to  materially  affect  the  shape  of  the  resultant 
wave. 

PROBLEMS 

1.  Draw  the  clock  and  sine  wave  diagrams  of  an  electro- 
motive force  whose  maximum  value  is  10  volts,  produced  by  a 
rectangular  coil  turning  in  a  uniform  magnetic  field  at  the  rate 
of  10  revolutions  per  second. 

2.  What  is  the  virtual  value  of  the  electro-motive  force  in 
Problem  1  ? 

SUGGESTION.  —  The   Vmean2  value  of  the  ordinate  of  a  sine  curve  is 

— —  of  the  maximum  ordinate. 
V2 


60  ELECTRICAL  PROBLEMS 

3.  Add  to  the  clock  and  sine  wave  diagrams  of  Problem  1  a 
vector  and  sine  wave  representing  a  current  with  a  maximum 
value  of  5  amperes  lagging  30°  in  phase  behind  the  electro- 
motive force. 

4.  What  is  the  virtual  value  of  the  current  in  Problem  3  ? 

5.  Draw  the  clock  and  sine  wave  diagrams  of  a  current  of 
21.21   amperes  lagging   50°   behind  an  electro-motive  force  of 
35.35  volts. 

NOTE.  —  In  speaking  of  volts  and  amperes,  unless  otherwise  specified, 
virtual  values  of  current  and  electro-motive  force  are  always  given. 

6.  Plot  the  curve  represented  by  e  =  100  sin  503  t. 

7.  Combine  with  the  curve  of  Problem  6  the  curve  repre- 
sented by  e  =  40  sin  (503  t  -  30°). 

SUGGESTION.  —  Evidently  the  zero  of  this  curve  will  be  displaced  J^,  of 
a  period  to  the  right  of  the  other  curve.  The  2  curves  are  combined  by 
adding  their  contemporaneous  ordinates. 

QUERY.  —  What  can  you  say  concerning  the  resultant  of  2  sine  curves 
of  the  same  periodicity  ? 

8.  Plot  the  curve  represented  by 

e  =  120  sin  80  irt  +  50  sin  400  irt, 

which  means  that  of  the  2  components  1  has  a  maximum  of 
120  volts  and  a  periodicity  of  40,  the  other  a  maximum  of  50 
volts  and  a  periodicity  of  200. 

9.  Plot  the  curve  represented  by 

i  =  50  sin  200  irt  +  15  sin  (600-7^-135°). 

10.  Plot  the  curve  represented  by 

e  =  50  sin  100  irt  +  10  sin  (500  irt  -  90°). 

11.  Plot  the  curve  represented  by 

i  =  100  sin  100  jrt  4- 15  sin  300  Trt  +  10  sin  (500  irt  -  180°). 

12.  Draw   the    clock    and  sine   wave   diagrams   of    the   two 
electro-motive  forces   of  a   2  phase  alternating  current   arma- 
ture, the   maximum  value   of  the  electro-motive  forces    being 
100  volts,  and  the  periodicity  60. 


ELECTRO-MOTIVE  FORCES   AND  CURRENTS        61 

SUGGESTION.  —  The  rotating  vectors  OA  and  OB  (Fig.  2)  may  represent 
the  maximum  values  of  the  electro-motive  forces  in  the  2  armature 
circuits,  Oa  being  the  instantaneous  value  of  the  electro-motive  force  in 
one  circuit  at  a  certain  instant,  and  Ob  the  contemporaneous  value  of  the 
electro-motive  force  in  the  other  circuit.  The  sine  waves  may  be  drawn 
separately,  each  with  reference  to  its  particular  vector. 

13.  Draw  the  clock  and  sine  wave  diagrams  of  the  3  electro- 
motive forces  of  a  3  phase  armature  with  star  grouping  of  the 


FIG.  2 


coils,  the  maximum  values  of  the  electro-motive  forces  being 
1000,  and  the  periodicity  25.  Also  draw  the  wave  diagrams  of 
the  electro-motive  forces  between  the  lines. 

When  the  wave  form  A  (Fig.  3)  of  a  periodic  function  is  not 
a  simple  sine  curve  the  virtual  values  are  less  easily  obtained, 
but  the  following  method  enables  them  to  be  obtained  graphi- 
cally. 

Rotate  a  vector  about  a  point,  laying  off  at  definite  intervals 
the  value  of  the  ordinate  of  the  wave  diagram  at  an  angular 
advance  from  the  beginning  of  the  cycle  equal  to  the  angle 
through  which  the  vector  has  rotated.  If,  as  is  usually  the 
case,  the  positive  and  negative  half  waves  are  equivalent,  only 
half  a  cycle  need  be  laid  off,  since  this  half  will  give  a  closed 


ELECTRICAL    PROBLEMS 


curve  as  shown  in  A  (Fig.  4),  which  is  the  polar  diagram  of  the 
wave  A  (Fig.  3).  With  a  planimeter  determine  the  area  of  the 
A 

B 


FIG.  3 


120 


00° 


140 


40' 


ISO 


0 
FIG.  4 


curve  and  construct  a  circle  B  of  equal  area,  tangent  to  the 
horizontal  axis  at  the  origin.     This  circle  is  the  polar  diagram 


ELECTRO-MOTIVE  FOECES  AND   CUEEENTS        63 


of  the  sine  curve  having  the  same  Vmean2  valne  of  the  equi- 
distant ordinates  as  the  original  curve.  Hence  the  diameter  of 
this  circle  is  the  maximum  ordiiiate  of  what  is  known  as  the 
equivalent  sine  curve,  B  (Fig.  3). 

14.  A  certain  electro-motive  force  is  the  resultant  of  2  simple 
harmonic  waves;   the  first  has  a  maximum  value  of  100  volts 
and  a  periodicity  of  50 ;   the  second  a  maximum  value  of  15 
volts  and  a  periodicity  of  150,  and    agrees  with  the  first   in 
phase  at  the  beginning  of  each  period  of  the  first.     Draw  the 
resultant  wave  form,  determine  the  virtual  value,  and  draw  the 
equivalent  sine  wave. 

15.  A  certain  current  wave  is  the  resultant  of  2  simple  sine 
waves;  the  fundamental  has  a  maximum  value  of  100  amperes 
and  a  periodicity  of  50  ;   the  harmonic  a  maximum  value  of 
30  amperes  and  a  periodicity  of  150,  and  lags  in  phase  90° 
behind  the  fundamental  at  the  beginning  of  each  period.    Draw 
the  resultant  wave  form  and  determine  the  virtual  value  of 
the  current. 


CHAPTER    XIV 
COMBINATION  OF  ALTERNATING  ELECTRO-MOTIVE  FORCES 

Two  harmonically  varying  electro-motive  forces  of  the  same 
periodicity  in  series  may  be  added  exactly  as  two  mechanical 
forces  are  added,  i.e.,  by  a  vector  or  clock  diagram. 


-x 


FIG.  5 

In  Fig.  5  let  OA  and  OB  represent  the  maximum  values  of 
El  and  Ey  the  two  electro-motive  forces  to  be  added,  6  the  angle 
of  phase  difference  between  El  and  E^. 

In  the  diagram  e1  =  El  sin  cot,  represented  by  Oa,  is  the 
instantaneous  value  of  the  first  electro-motive  force,  and 
ez  =  E^  sin  (cot  —  6),  represented  by  Ob,  is  the  contemporaneous 
value  of  the  second  electro-motive  force. 

The  vector  sum  of  OA  and  OB  is  evidently  OC,  and  the 
instantaneous  value  of  OC  is  Oc.  Now  whatever  the  position 

64 


ALTERNATING  ELECTEO-MOTIVE  FOKCES          65 

of  the  parallelogram  AOBC  when  rotated  about  the  center  0, 
the  sum  of  the  projections  Oa  and  Ob  must  always  be  equal  to 
Oc.  Evidently,  then,  the  sum  of  the  two  harmonically  varying 
electro-motive  forces  of  the  same  periodicity  is  itself  a  harmon- 
ically varying  electro-motive  force  of  the  same  periodicity  as  its 
components  and  intermediate  in  phase  between  them. 
From  the  trigonometry  of  the  figure  we  have 

6>~6'2  =  6U2  +  OZf  +  2  OA  X  OB  cos  0 

~OC2  -  oZa  -  OB2 
and  cos  6  = 


2  x  OA  x  OB 

We  have  already  seen  that  with  harmonically  varying  quan- 
tities virtual  values  have  a  fixed  relation  to  maximum.  The 
vector  diagram  therefore  is  as  applicable  to  the  composition  of 
virtual  as  of  maximum  values. 

PROBLEMS 

1.  Two  electro-motive  forces  of  the  same  periodicity  and  of 
80  and  60  volts,  respectively,  differing  in  phase  by  37°,  act  in 
series.     What  is  the  resultant  voltage  ? 

2.  The  electro-motive  forces  of  Problem  1  differ  in  phase  by 
30°.     Find  their  resultant. 

3.  The  electro-motive  forces  of  Problem  1  differ  in  phase  by 
60°.     Find  their  resultant. 

4.  Two  electro-motive  forces  of  the  same  periodicity  and  each 
of  100  volts  differ  in  phase  successively  by  30°,  60°,  90°,  120°, 
150°,  and  180°.     Find  the  resultant  voltage  in  each  case. 

5.  Two  electro-motive  forces  of  the  same  periodicity  and  of 
40  and  30  volts,  respectively,  are  in  series.      Their  resultant  is 
50  volts.     What  is  their  phase  difference? 

6.  If  the  40  volt  component  of  Problem  5  leads,  by  what 
angle  does  the  resultant  lag  behind  it  ? 

7.  If   the    2    electro-motive    forces    of    Problem    5    have   a 
resultant  of  60  volts,  by  what  angle  do  they  differ  in  phase  ? 
By  what  angle  does  the  40  volt  component  lead  the  resultant  ? 


66 


ELECTRICAL  PROBLEMS 


8.  Two  equal  period  electro-motive  forces  of  100  and   25 
volts,  respectively,  are  in  quadrature.     What  is  the  resultant? 

9.  Two   electro-motive   forces   are   in   quadrature.      One  is 
1000  volts  and  their  resultant  is  1010  volts.     What  is  the  sec- 
ond component  and  what  is  its  phase  relation  to  the  resultant? 


10.  Four  equal  period  electro-motive  forces  of  30,  40,  50,  and 
60  volts,  respectively,  are  in  series  ;  they  successively  differ  in 
phase  by  30°.  Find  the  resultant  in  magnitude  and  phase. 

SUGGESTION.  —  The  4  above  electro-motive  forces  may  be  combined  in 
pairs  by  the  vector  diagram,  and  the  2  resultants  thus  obtained  combined 
for  the  final  resultant.  Where  there  are  more  than  2  components,  however, 
it  is  usually  preferable  to  combine  them  by  the  method  of  rectilinear  coordi- 
nates ;  that  is,  resolve  each  electro-motive  force  into  2  components  along 


ALTERNATING  ELECTRO-MOTIVE  FORCES          67 

perpendicular  axes  ;  add  all  the  X  components  thus  obtained  and  all  the  Y 
components  ;  their  sums  are  the  A"  and  Y  components  of  the  resultant  electro- 
motive force.  Thus  in  Fig.  6  we  may  lay  out  the  30  volt  component  OA  on 
the  X  axis  and  the  60  volt  component  OD  on  the  Y  axis,  with  the  40  volt 
and  the  50  volt  components  OB  and  OC  between  at  the  proper  angles  ;  then 
denoting  by  OAX  and  OAy  the  X  and  Y  components  of  OA,  etc.,  we  have 
OAX  =  30  cos  0°  =  30  OAy  =  30  sin  0°  =  0 

OBX  =  40  cos  30°  =  34.64  OBy  =  40  sin  30°  =  20 

OCX  =  50  cos  60°  =  25  OCy  =  50  sin  60°  =  43.3 

ODX  =  60  cos  90°  =    0  ODy  =60  sin  90°  =  60 

OEX  =  89.64  OEy  =  123.3 

The  resultant  OE  of  OEX  and   OEy  in  quadrature  is  152.4.     This  is 
therefore  the  resultant  of  all  4  original  components.      OE  is  in  advance  of 

1°3  3 
OA  by  tan-  ^  =  63°  59'. 

11.  The  same  electro-motive  forces  as  in  Problem  10  have 
phase  angles  of  0°,  60°,  120°,  and  180°,  respectively.     Find  the 
value  and  phase  angle  of  the  resultant. 

12.  Three  electro-motive  forces,  each   of  100  volts,  are  in 
series  with  phase  angles  of  30°,  150°,  and  270°.    What  is  their 
resultant  in  magnitude  ? 

13.  A  3  phase  star  connected  armature  generates  1000  volts 
in  each  armature  circuit.     The  phase  angles  of  the  3  electro- 


FIG.  7 

motive  forces  being  respectively  0°,  120°,  and  240°,  what  will 
be  the  magnitude  and  phase  angles  of  the  electro-motive  forces 
between  the  line  wires  ? 

SUGGESTION.  —  In  Fig.  7,  OA,  OB,  and  OC  are  the  vectors  representing 
the  electro-motive  forces  in  the  3  armature  circuits.  A  A',  BE',  and  CC' 
are  the  3  line  wires,  and  A'B',  B'C',  and  CM"  are  the  vectors  representing 
the  3  line  voltages. 


68  ELECTRICAL  PROBLEMS 

14.  Power  is  transmitted  over  3   wires  from  a   1000   volt 
2  phase  alternator,  2  of  its  4  armature  leads  being  connected 
at  the   brushes.     Calling  the   phase  angles  of  the  2  armature 
electro-motive  forces  0°  and  90°,  what  in  magnitude  and  phase 
are  the  3  line  voltages  ? 

15.  Power  is  transmitted  over  4  wires  from  a  4  phase  alter- 
nator made  by  connecting  the  2  armature  circuits  of  a  1000 
volt  2  phase  alternator  at  their  middle  points.     What  are  the 
magnitudes  of  the  4  line  voltages  ? 


CHAPTER    XV 
COMBINATION   OF  ALTERNATING   CURRENTS 

Two  alternating  currents  uniting  in  a  common  conductor  are 
combined  precisely  as  two  alternating  electro-motive  forces  in 
series. 

Thus  in  Fig.  8  let  there  be  a  harmonic  current  of  30  amperes 
in  A  and  a  harmonic  current  of  40  amperes  in  B,  the  tAvo  currents 
differing  90°  in  phase.  Then, by  the  vector  diagram  (Fig.  9),  there 
will  be  a  current  of  50  amperes  in  C ;  for,  since  the  point  of  junc- 
tion 0  is  without  capacity,  it  is  evident  that  the  instantaneous 
value  of  C  is  the  sum  of  the  instantaneous  values  of  A  and  B. 

Accordingly,  as  in  Fig.  10,  C  may  be  plotted  as  the  algebraic 
sum  of  the  contemporaneous  ordinates  of  the  sine  curves  A  and 
B,  differing  in  phase  by  90°.  In  the  problems  of  this  chapter 
sine  waves  of  current  are  assumed  unless  otherwise  stated. 

PROBLEMS 

1.  In  a  quarter  phase  system  of  2  equal  currents  what  is  the 
ratio  of  each  outgoing  current  to  the  current  in  the  common 
return  ? 

NOTE.  —  A  quarter  phase  system  has  two  currents  with  a  phase  difference 
of  90°,  i.e.,  in  quadrature. 

2.  What  is  the  resultant  of  2  currents,  each  of  40  amperes, 
differing  120°  in  phase? 

3.  In  a  3  phase  system  with  3  equal  currents,  differing  in  phase 
by  120°,  what  is  the  value  of  the  current  in  the  common  return  ? 

NOTE. It  thus  appears  that  in  a  balanced  symmetrical  3  phase  system 

a  fourth  conductor  serves  no  purpose  whatever. 

69 


ELECTRICAL  PROBLEMS 


4.  An  unbalanced  quarter  phase  system  has  currents  of  12 
amperes  and  5  amperes,  respectively,  in  its  branches.  What  is  the 
magnitude  and  phase  angle  of  the  current  in  the  common  return  ? 

At .      c 

\ 
| 

/'          i 


X 


X 


x 


X 


X 


X 


X 


X 


X 


40 


FIG.  8 


FIG.  9 


75  - 


.",0 


25-  - 


8,0 


75 -1- 


/   / 


/ 


70T7X  42.42  =  30 

707X56.56=40 

.707  X  70.7=50 


FIG.  10 


5.  An  unbalanced  3  phase  system  has  currents  in  the  3  legs 
of  4,  5,  and  7  amperes,  respectively,  differing  in  phase  by  120°. 
What  is  the  current  in  a  fourth  conductor  serving  as  a  common 
return  ? 

SUGGESTION. — Solve  by  the  method  of  rectangular  coordinates,  as 
described  in  Chapter  XIV,  page  66. 


COMBINATION  OF  ALTERNATING  CURRENTS       71 

6.  A  current  divides  between  2  branches ;  after  division  it 
is  found  that  one  branch  carries  10  amperes,  the  other  6  amperes, 
and   the    phase    difference   between   the  two    currents   is   30°. 
What  was  the  current  before  division  in  magnitude  and  phase  ? 

SUGGESTION.  —  This  mafy  be  solved  graphically,  as  shown  in  Fig.  11. 
Draw  OA  and  OB  making  angle  A  OB  =  30°;  lay  off  OC  and  OD  in  the 
ratio  of  6:10.  Complete  the  parallelogram  OCED  with  OC  and  OD  as 
adjacent  sides,  and  draw  the  diagonal  OE.  OE  will  represent  the  current 
before  division  in  magnitude  and  phase  on  the  scale  of  OA  and  OB. 
Angle  EOD  or  EOC,  the  phase  angle  between  the  original  current  and  a 
component,  may  be  found  with  a  protractor  or  by  trigonometry. 

7.  Two  generators  in  parallel  supply  a  current  of  25  amperes 
to  a  line.     The  currents  in  the  2  armatures  are  respectively 


16.*9  - 


I) 
FIG.  11 

10°  ahead  of  and  25°  behind  the  main  current  in  phase.     What 
is  the  magnitude  of  the  current  in  each  armature  ? 

8.  Two  currents  of  10  amperes  each,  differing  160°  in  phase, 
unite  in  a  common  return.     Find  the  magnitude  and  phase  of 
the  return  current. 

9.  Currents  of  5,  6,  7,  and  8  amperes,  having  phase  angles  of 
0°,  30°,  90°,  and  150°,  respectively,  unite  in  a  common  conductor. 
What  is  the  resultant  current  in  magnitude  and  phase  ? 

10.  Currents  of  10,  20,  30,  40,  50,  60,  70,  and  80  amperes, 
with  phase  angles  of  0°,  45°,  90°,  135°,  180°,  225°,  270°,  and 
315°,  respectively,  unite  in  a  common  conductor.     What  is  the 
resultant  current  in  magnitude  and  phase  ? 

11.  Currents  of  20,  35,  10,  90,  3,  12,  47,  20,  and  30  amperes 
have  phase  angles  of  30°,  135°,  330°,  60°,  180°,  210°,  90°,  300°, 


72  ELECTRICAL  PROBLEMS 

and   45°,   respectively.     They   unite   in   a   common  conductor. 
What  is  the  resultant  current  in  magnitude  and  phase  ? 

12.  A  symmetrical  3  phase  armature  with  delta  grouping  of 
armature  circuits  generates  100  amperes  in  each  circuit.    What 
is  the  current  in  each  of  the  line  wires  ? 

13.  A  condenser  and  kicking  coil  are  placed  in  multiple  in 
an  alternating  current  circuit.     The  condenser  has  a  current 
of  1  ampere  85°  in  advance  of  the  main  current,  and  the  kicking 
coil  a  current  of  1  ampere  85°  behind  the  main  current  in  phase. 
What  is  the  value  of  the  main  current  ? 


CHAPTER    XVI 
IMPEDANCE 

The  electro-motive  force  required  to  establish  an  alternating 
current  through  a  simple  resistance,  or  the  ohmic  drop  in  that 
resistance,  is,  as  is  the  case  with  direct  currents,  the  product  of 
the  current  and  resistance,  E  —  IR. 

All  circuits,  however,  possess  more  or  less  inductance  and 
capacity,  both  of  which,  when  the  circuit  is  traversed  by  an 
alternating  current,  give  rise  to  reactive  electro-motive  forces  in 
quadrature  with  the  current.  The  electro-motive  force  due  to 
inductance  is  90°  behind  the  current  in  phase,  and  is  numerically 


I) 

^1 

r                                     ^-"^' 

.J^""" 

io>J 

i  ^"lT  *i 

R 

Loll 


FIG.  12  FIG.  13 

equal  to  Lai;  that  due  to  capacity  is  90°  ahead  of  the  current  in 

phase,  and  is  expressed  by  -=-;  co  is  of  the  nature  of  an  angular 

C/  co 

velocity,  and  is  equal  to  2  ?rn,  n  being  the  periodicity. 

Thus  in  Fig.  12  let  there  be  an  alternating  current  of  virtual 
value  I  and  periodicity  n  in  a  circuit  of  resistance  R  and 
inductance  L.  Then  OB  =  RI  represents  the  effective  electro- 
motive force,  or  that  required  to  overcome  the  resistance,  and 
OC=Lo)Ithe  reactive  electro-motive  force  due  to  inductance, 

73 


74  ELECTRICAL  PROBLEMS 

90°  behind  HI  in  phase.  To  overcome  OC  there  must  be 
impressed  on  the  circuit  an  equal  and  opposite  electro-motive 
force  OZ>,  which  is  accordingly  90°  ahead  of  OB.  The  total 
impressed  electro-motive  force  E  must  therefore  be  the  resultant 
or  vector  sum,  OA,  of  OB  and  OD,  equal  to  /  V/t>2  +  L'2co*. 

If  we  divide  each  side  of  the  triangle  of  electro-motive  forces 
by  the  current  /,  we  shall  have  the  ohmic  triangle  OAB  (Fig.  13), 
in  which  the  base  represents  the  resistance  R,  the  perpendic- 
ular the  reactance  .£&>,  and  the  hypoteneuse  the  impedance 
V  ffi  -f-  Z,2o>2,  all  expressible  in  ohms. 

It  is  evident  that  the  current  will  lag  behind  the  electro- 
motive force  by  an  angle  0,  and  that 

Leo  Leo  R 

tan  6  =  —  ,  sm  6  =  -—===,  cos  6  = 
R 


and  that  the  current 

E  E 


sin  (cot  -  0), 

where  i  is  the  instantaneous  value  of  the  current,  and  cot  the 
contemporaneous  phase  angle  of  the  electro-motive  force. 

PROBLEMS 

1.  What  is  the  impedance  in  a  circuit  of  4  ohms  resistance 
and  3  ohms  reactance  ? 

Solution.     Impedance  =  ^ Rz  +  La)2  =  V42  +  32  =  5  ohms. 

2.  What  is  the  inductance  in   Problem  1  if  the  frequency 
is  159  ? 

3.  By  what  angle  does  the  current  lag  behind  the  electro- 
motive force  in  the  circuit  of  Problem  1  ? 

4.  Assuming  the  virtual  value  of  the  electro-motive  force  in 
Problem  1  to  be  100  volts,  draw  the  sine  curves  of  the  various 
electro-motive  forces  and  current. 


IMPEDANCE  75 

5.  With  a  frequency  of  50,  what  is  the  impedance  of  a  coil 
of  5  ohms  resistance  and  .01  henry  inductance  ? 

6.  With  an  impressed  electro-motive  force  of  100  volts  and  a 
frequency  of  60  cycles  per  second,  what  current  will  be  produced 
in  a  circuit  of  4  ohms  resistance  and  .02  henry  inductance? 

7.  A  current  of  18.48  amperes,  with  a  periodicity  of  33,  flows 
in  a  coil  of  10  ohms  resistance  and  .02  henry  inductance.    What 
is  the  impressed  electro-motive  force  ? 

8.  What  must  be  the  periodicity  in  Problem  7  to  give  a 
current  of  8  amperes? 

9.  By  what  angles  will  the  current  lag  behind  the  electro- 
motive force  in  Problems  7  and  8  ? 

10.  With  a  periodicity  of  25  cycles  per  second,  the  impedance 
of  a  circuit  of  12  ohms  resistance  is  20  ohms.     What  is  the 
inductance  ? 

11.  With  25  cycles  per  second,  the  impedance  of  a  circuit  of 
.05  henry  inductance  is  20  ohms.     What  is  the  resistance? 

12.  Find  analytically  and  graphically  the   impedance  of  a 
circuit  of  200  ohms  resistance'  and  .4  henry  inductance,  with 
a  periodicity  of  70. 

SUGGESTION.  —  Draw  ohmic   triangle  as  in  Fig.  13  and  measure  the 
hypotenuse. 

13.  Find  graphically  the  reactance  of  a  circuit  whose  resist- 
ance is  4  ohms  and  whose  impedance  is  10  ohms. 

14.  Two  No.  1  B.S.G.  wires  18  inches  apart  extend  as  feeders 
from  an  electric  light  station  to  a  center  of  distribution  3  miles 
away.     Neglecting    capacity,   what    is    the    impedance    of   the 
feeders  at  a  temperature  of  60°  F.  and  a  periodicity  of  125  ? 

15.  A  single  phase  transmission  line  40  miles^long  consists 
of  2  No.  0  B.S.G.  wires  30  inches  apart.     Neglecting  capacity, 
what  is  the  impedance  of  the  line  at  the  temperature  of  0°  C., 
with  a  periodicity  of  25  ? 

16.  A  circuit  consists  of  2  sections  in  series,  one  of  12  ohms 
resistance  and  .015   henry  inductance,  the  other  of    10  ohms 


76 


ELECTRICAL  PROBLEMS 


resistance    and   .05    henry   inductance, 
impedance  at  a  frequency  of  40. 


Find   graphically   the 


SUGGESTION.  —  2  irn  =  <o  =  251.3.  Draw  OA  (Fig.  14)  =  12,  AB  per- 
pendicular to  it  =  .015  o).  Then  OB  =  Vl22  +  (.015  o>)2  =  impedance  of 
the  first  section.  Draw  EC  parallel  to  OE  =  10,  CD  perpendicular  to 
it  =  .05  w.  Then  BD  =  Vlo2  +  (.05  o>)2  =  impedance  of  the  second  sec- 
tion. Draw  OD  =  V(10  +  12)2  +  (.015  +  .05)2a>2  =  impedance  of  the  circuit. 

NOTE.  —  From  an  inspection  of  Fig.  14  it  is  evident  that  two  or  more 
reactances,  and  therefore  two  or  more  inductances,  in  series  may  be  com- 
bined by  simple  addition,  as  is  the  case  with  resistances.  The  resultant 
impedance  of  two  or  more  impedances  in  series  is,  however,  the  vector  sum. 


FIG.  14 


17.  Two  coils  are  connected  in  series ;  the  first  has  a  resist- 
ance of  12  ohms  and  an  inductance  of  .12  henry,  the  second  a 
resistance  of  26  ohms  and  an  inductance  of  .06  henry.     What 
must  be  the  periodicity  of  an  electro-motive  force  of  400  volts 
that  produces  a  current  of  5  amperes  through  these  coils  ? 

18.  An    electro-motive   force   with   a   periodicity   of   100   is 
impressed  upon  a  circuit  with  a  resistance  of  200  ohms.     Plot 
a  curve  of  impedance  as  the  inductance  varies  from  0  to  .4  henry. 


IMPEDANCE  77 

19.  An   electro-motive   force   with   a   periodicity   of    100   is 
impressed  upon  a  circuit  with  an  inductance  of  .2  henry.     Plot 
a  curve  of  impedance  as  the  resistance  varies  from  0  to  200 
ohms. 

20.  Plot  2  current  curves :  one  when  an  electro-motive  force 
of  2000  volts  with  a  periodicity  of  80  acts  upon  a  circuit  with 
a   constant   inductance   of  .5  henry  and   a  resistance    varying 
from  0  to  200  ohms ;    the  other  when  the  same  electro-motive 
force  acts  upon  a  circuit  with  a  constant  resistance  of  50  ohms 
and  an  inductance  varying  from  0  to  .5  henry. 

Fig.  15  represents  the  electro-motive  force  triangle  of  a 
circuit  of  resistance  R  and  capacity  C  acted  upon  by  an  electro- 
motive force  of  periodicity  n  = OB  is  the  effective  electro- 

2  7T 

motive  force  in  phase  with  the  current  and  numerically  equal 
c 

cu3 

RI  '^   B  R 


I 


FIG.  15  FIG.  16 

to  RI;  OC  is  the  reactive  electro-motive  force  90°  ahead  of  the 
current  in  phase  and  equal  to  — •  •>  while  OD  =  BA  is  that  com- 
ponent of  the  impressed  electro-motive  force  required  to  over- 
come OC.  The  total  impressed  electro-motive  force  is  accordingly 

the  resultant  OA  of  OB  and  OD  and  equals  J\/^2  +  -^- 
Evidently  the  current  will  lead  the  impressed  electro-motive 

force  in  phase  by  an  angle   0  whose  tangent  is ^  RI,  or 

1  Ceo 

0-tan-1— • 
Crco 


78  ELECTRICAL   PROBLEMS 

In  the  same  way  as  in  the  inductive  circuit  the  ohmic  triangle 
is  as  shown  in  Fig.  16,  excepting  that  6  is  now  the  angle  of 

V/  i  \2 
E2  +  f  —  )  • 
\C(oJ 

21.  What  is  the  impedance  in  a  circuit  of  12  ohms  resistance 
and  5  ohms  capacity  reactance  ? 

22.  What  is  the  capacity  in  Problem  21  if  the  periodicity  is 
125? 

23.  If  an  electro-motive  force  of  100  volts  is  impressed  on 
the  circuit  of  Problem  21,  what  current  is  produced? 

24.  By  what  angle  does  the  current  lead  the  electro-motive 
force  in  Problem  23  ? 

25.  Draw  the  sine  curves  of  the  various  electro-motive  forces 
and  current  in  the  case  of  Problems  21  to  24. 

26.  When  the  periodicity  is  100,  what  is  the  impedance  of  a 
line  containing  200  ohms  resistance  and  5  microfarads  capacity? 

27.  In  Problem  26  what  capacity  will  double  the  impedance  ? 

28.  With  a  periodicity  of  54,  what  is  the  impedance  of  a 
circuit  of  5000  ohms  resistance  and  1  microfarad  capacity  ? 

29.  If  the  resistance  in  Problem  28  is  doubled,  how  and  by 
what  per  cent  will  the  impedance  be  affected  ? 

30.  If  the  resistance  of  Problem  28  is  reduced  one  half,  how 
and  by  what  per  cent  will  the  impedance  be  affected  ? 

31.  If  the  capacity  in  Problem  28  is  doubled,  how  and  by 
what  per  cent  will  the  impedance  be  affected? 

32.  If  the  capacity  in  Problem  28  is  reduced  one  half,  how 
and  by  what  per  cent  will  the  impedance  be  affected  ? 

33.  If  the  frequency  in  Problem  28  be  changed  to  34,  what 
value  must  be  given  to  the  resistance,  the  capacity  remaining 
1  microfarad,  to  keep  the  impedance  at  5804  ohms? 

34.  With  a  periodicity  of  100,  find  graphically  the  impedance 
of  a  circuit  of  175  ohms  resistance  and  5  microfarads  capacity. 

35.  Find  graphically  the  resistance  of  a  circuit  of  13  micro- 
farads capacity  whose  impedance  is  216  ohms  at  a  frequency 
of  72. 


IMPEDANCE 


79 


36.  In  a  circuit  of  4  microfarads  capacity  the  periodicity  is 
80.     Draw  a  curve  of  impedance  as  the  resistance  varies  from 
0  to  1200  ohms. 

37.  In  a  circuit  of  600  ohms  resistance  the  periodicity  is  80. 
Draw  a  curve  of  impedance  as  the  capacity  varies  from  2  to  8 
microfarads. 

38.  A  constant  difference  of  potential  of  2000  volts  at  a  fre- 
quency of  125  is  maintained  between  the  terminals  of  a  circuit. 
Draw  2  curves  of  current:  one  with  a  constant  resistance  of 
100  ohms  and  a  capacity  varying  from  0  to  100  microfarads  ; 
the   other  with  a  constant  capacity  of  10   microfarads   and   a 
resistance  varying  from  0  to  1000  ohms. 

When  both  inductive  and  capacity  reactance  electro-motive 
forces  are  large  enough  to  be  taken  into  account,  one,  due  to 


FIG.  17 


FIG.  18 


inductance  =  Lcol,  will  be  90°  behind  the  current  in  phase ;  the 

other,  due  to  capacity  =  — -  •>  will  be  90°  ahead  of  the  current 

,  Ceo 

in  phase. 

These  are  shown  by  OB  and  0(7,  respectively,  in  Fig.  17, 
where  OA  represents  the  phase  relations  of  the  current  and  the 
magnitude  of  the  ohmic  drop.  The  algebraic  sum  of  the  two 

reactance   electro-motive    forces,    Lcol  and  -—->   is  the  net  or 

Ceo 

effective  reactance  ejectro-motive  force  represented  by  OD.     It 

is  clear  that  either  Leo  or  --  may  be  the  greater,  the  net  react- 

C/co 

ance  accordingly  being  either  90°  behind  or  90°  ahead  of  the 


80  ELECTRICAL  PROBLEMS 

current  in  phase.  To  overcome  the  reactance  there  must  be 
impressed  upon  the  circuit  an  electro-motive  force  AE,  equal 
and  opposite  to  the  net  reactive  electro-motive  force  OD,  while 
to  overcome  the  resistance  there  must  be  an  electro-motive  force 
El  in  phase  with  the  current ;  hence  the  total  impressed  electro- 
motive force  is  represented  in  magnitude  and  phase  by  OE,  the 

r~.  ~T~     i  \  2 

resultant  of  OA  and  AE,  equal  to  /  \  A>2  -f  (  Leo —  )  • 

Fig.  18  shows  the  corresponding  ohmic  triangle.     The  tangent 

of  the  angle  of  lag  or  lead  is  evidently — ,  the  current 

lagging  or  leading  according  as  LCD  or  —  is  the  greater. 

The  impedance  of  a  circuit  containing  inductive  and  capacity 

r~>  /      i  \2 

reactance  is  expressed  by  \  R*  +  (  Leo  -      -  )  .     Plainly,  where 


Leo  =  -—  >    or    where    LCco2  =  1,    the    impedance    equals    the 
(7&) 

resistance. 

39.  What  is  the  impedance  of  a  circuit  of  14  ohms  inductive 
reactance,  5  ohms  capacity  reactance,  and  12  ohms  resistance  ? 

40.  With  an   impressed  electro-motive   force   of    100   volts, 
what  current  will  flow  in  the  circuit  of  Problem  39  ? 

41.  Does  the  current  of  Problem  40  lag  or  lead,  and  by  what 
angle  ? 

42.  Draw  the  sine  curves  of  the  various  electro-motive  forces 
and  current  in  Problems  39  to  41. 

43.  With  a  periodicity  of  200,  what  is  the  impedance  of  a 
circuit  of  10  ohms  resistance,  .3  henry  inductance,  and  2  micro- 
farads capacity? 

44.  With  a  periodicity  of   80,  what  is  the  impedance  of  a 
circuit   of    180    ohms   resistance,    5    henrys    inductance,    and 
.76  microfarad  capacity? 

45.  What  are  the  phase  relations  of  the  current  and  electro- 
motive force  in  the  circuit  of  Problem  44  ? 


IMPEDANCE  81 

46.  All  other  conditions  in  Problem  44  remaining  the  same, 
what  value  of  L  will  make  the  impedance  180  ohms  ? 

47.  What  current  will  flow  in  a  circuit  of  5  ohms  resistance, 
.01  henry  inductance,  and  100  microfarads  capacity,  if  1200  volts 
with  a  frequency  of  120  are  impressed  upon  its  terminals? 

48.  In  Problem  47  will  the  current  lag  or  lead,  and  by  what 
angle  ? 

49.  A  circuit  in  which  the  frequency  is  50  contains  500  ohms 
resistance  and  1  henry  inductance.    Draw  a  curve  of  the  impe- 
dance as  the  capacity  varies  from  0  to  20  microfarads. 

50.  A  circuit  in  which  the  frequency  is  50  contains  500  ohms 
resistance  and  10  microfarads  capacity.     Draw  a  curve  of  the 
impedance  as  the  inductance  varies  from  0  to  5  henrys. 

51.  A  circuit  in  which  the  frequency  is  50  contains  25  henrys 
inductance  and  10  microfarads  capacity.    Draw  a  curve  of  the 
impedance  as  the  resistance  varies  from  0  to  1000  ohms. 

52.  An  electro-motive  force  of  2300  volts  is  impressed  upon 
a  circuit.    Draw  4  curves  of  current :  the  first  with  100  ohms 
resistance,   1    henry   inductance,   a  frequency  of    25,  and  the 
capacity  varying  from  0  to  100  microfarads;   the  second  with 
100  ohms  resistance,  25  microfarads  capacity,  a  frequency  of  25, 
and  inductance  varying  from  0  to  10  henrys  ;  the  third  with 
1  henry  inductance,  25  microfarads  capacity,  a  frequency  of  25, 
and  resistance  varying  from  0  to  1000  ohms;  and  the  fourth 
with  100  ohms  resistance,  1  henry  inductance,  25  microfarads 
capacity,  and  the  frequency  varying  from  0  to  50. 

53.  An  electro-motive  force  of  100  volts,  with  a  periodicity 
of  80,  is    impressed   upon   a   circuit  of    10    ohms    resistance, 
20    microfarads    capacity,    and    .2    henry   inductance,    all    in 
series.     What  is  the   voltage   between   the    terminals    of   the 
condenser,    and    what    that    between    the    extremities    of    the 
inductance  ? 

54.  Draw  the  voltage  curves  at  the  terminals  of  the  capacity 
and  of  the  inductance  under  the  conditions  of  Problem  53  as 
the  capacity  varies  from  0  to  100  microfarads. 


82 


ELECTRICAL   PROBLEMS 


55.  Draw  the   above   curves   under  the  conditions  of  Prob- 
lem 53  as  the  inductance  varies  from  0  to  1  henry. 

56.  Draw  the  above  curves  under  the  conditions   of  Prob- 
lem 53  as  the  resistance  varies  from  0  to  1000  ohms. 

57.  Draw  the  above  curves  under  the  conditions  of    Prob- 
lem 53  as  the  periodicity  varies  from  0  to  400  cycles  per  second. 

When  a  circuit  divides  into  several  parallel  branches,  each 
possessing  resistance  and  reactance,  it  is  evident  that  there 
might  be  substituted  for  the  multiple  impedances 


+ 


etc.,    an    equivalent    impedance 


1 

CM 


1  ->    without 


change  of  the  main  current,  either  in  magnitude  or  phase. 

The  joint  impedance  of  parallel  branches  may  be  quite  easily 
obtained  in  the  following  manner : 

In  Fig.  19  let  .AOB,  AOC,  and  A 01)  represent  the  electro- 
motive force  triangles  of  3  parallel  branches  in  which  flow 


Fio.  1!) 

currents  Iv  /2,  and  J3.     Rv  /£2,  and  R3  are  the  resistances  of 
the  3  branches.     The  first  has  a  net  inductive  reactance  =  ZjO), 

the   second    a    net    capacity  reactance  = »   and  the  third  a 

net  inductive  reactance  =  L3co.     We  then  have  the  impressed 


IMPEDANCE  83 

electro-motive  force  represented  by  OA,   the  effective  electro- 
motive forces  .#!/!,  ^2/2,  and  ^Vs  represented  by  OB,  OC,  and  01), 

and  the  reactive  electro-motive  forces  L^wl^,    -^-  »    and  L3col? 

C    ® 


represented  by  AB,  AC,  and  ^4Z>.  Dividing  OB  by  .#r  0(7  by  ,K2, 
and  OD  by  ./£3,  we  get  the  separate  currents  represented  by 
Ob  =  Iv  Oc  —  Iy  and  Od  =  J3.  The  vector  sum  Oe  of  Iv  /2, 
and  J3  is  /,  the  current  in  the  main  line  before  branching,  and 
the  current  that  would  flow  if  a  single  conductor  of  equivalent 
impedance  was  substituted  for  the  parallel  branches.  OF  =  RI 

is  the  effective,  and  AF  '=  ll  La)  --  )  the  reactive,  component 

V  G»/        QF 

of  the  impressed  electro-motive  force  E.  —  =  -K  is  the  resist- 
ance. and  -  —  =  (  Lw  --  )  is  the  reactance  of  the  equivalent 


circuit.  The  phase  relations  of  the  various  currents  may  be 
obtained  either  analytically  by  the  method  of  Chapter  XIV 
or  with  a  protractor  from  the  graphical  diagram.  This  method 
is  of  course  applicable  to  a  circuit  of  any  number  of  branches, 
each  containing  either  inductance  or  capacity,  or  both. 

58.  An  alternating  current  of  25  amperes,  with  a  periodicity 
of  150,  flows  in  a  circuit  closed  in  one  place  by  parallel  branches, 
one  of  4  ohms  resistance  and  .003  henry  inductance,  the  other 
of  12  ohms  resistance  and  200  microfarads  capacity.     Determine 
the  impressed  electro-motive  force,  the  equivalent  impedance  of 
the  parallel  branches,  the  phase  relations  and  magnitude  of  the 
component  currents,  the  resistance  and  reactance  of  the  equiva- 
lent circuit,  and  the  phase  relation  of  the  main  current. 

59.  Two   branches,   one  of  4  microfarads  capacity  and  500 
ohms    resistance,   and    the    other    of    1    henry  inductance  and 
500  ohms  resistance,  have  impressed  upon  their  terminals  an 
alternating  electro-motive  force  of  1000  volts,  with  a  periodicity 
of  79.58  cycles  per  second.     What  is  the  equivalent  resistance, 
reactance,  and    impedance  of  the  circuit,  and  what  the  main 
current  in  magnitude  and  phase  ? 


84  ELECTRICAL  PROBLEMS 

60.  An  alternating  current  of  2  amperes  flows  in  a  circuit  of 
500  ohms  resistance,  the  frequency  being  79.58.     Draw  curves 
representing  the  varying  values  of  the  difference  of  potential 
at  the  terminals  of  the  circuit,  and  varying  angle  of  lead  as  the 
capacity  varies  from  1  to  16  microfarads. 

61.  An  electro-motive  force  of  11,000  volts,  with  a  perio- 
dicity of  50  cycles  per  second,  is  impressed  upon  a  circuit  of 
4  parallel  branches  :   one  of  200  ohms  resistance  and  1  henry 
inductance;  a  second  of  500  ohms  resistance  and  10  microfarads 
capacity;  a  third  of  250  ohms  resistance,  3  henrys  inductance, 
and  6  microfarads  capacity ;  and  the  fourth  of  600  ohms  resist- 
ance, 1  henry  inductance,  and  2  microfarads  capacity.     What 
current   will  flow   in    each  branch,    what   are    the    resistance, 
reactance,  and  impedance  of  the  equivalent  circuit,  and  what 
are  the  magnitude  and  phase  relations  of  the  main  current  ? 

62.  Alternating  currents  of  1  ampere  each,  with  a  periodicity 
of  39.79,  flow  in  2  parallel  branches,  one  of  40   microfarads 
capacity   and  10  ohms  resistance,  and  the  other  of  .4  henry 
inductance  and  10  ohms  resistance.    Determine  the  main  current 
in  phase  and  magnitude,  the   impressed    electro-motive  force, 
and   the   resistance,   reactance,   and   impedance  of   the   circuit 
equivalent  to  the  combination. 


CHAPTER    XVII 
DIRECT   CURRENT   ARMATURES 

The  fundamental  equation  of  the  continuous  current  dynamo 
is  E  =  2  nN^'LQ  ~8,  where  E  is  the  electro-motive  force  in  volts 
generated  in  one  circuit  of  the  armature  ;  n,  which  may  be  called 
the  periodicity,  is  the  number  of  revolutions  per  second  mul- 
tiplied by  the  number  of  pairs  of  pole  pieces  ;  N  is  the  number 
of  conductors  in  series  on  the  armature,  counted  from  positive 
to  negative  brush,  and  <I>  is  the  magnetic  flux  in  maxwells  or 
C.G.S.  lines  of  force  from  each  pole  piece  that  are  effectively 
cut  by  the  moving  conductors.  For  practical  purposes  it  may 
be  convenient  to  express  the  magnetic  flux  in  megamaxwells  or 
millions  of  maxwells,  in  which  case  the  fundamental  equation 
assumes  the  form  E  =  .02 


PROBLEMS 

1.  What    electro-motive    force    is    generated    by  a   4   pole 
continuous   current   dynamo   having    160    conductors    upon    a 
parallel  wound  drum  armature,  5,000,000  maxwells  from  each 
pole  piece  effectively  cut  by  the  moving  conductors,  and  a  speed 
of  900  revolutions  per  minute? 

Solution.  One  hundred  and  sixty  conductors  upon  the  face  of  the  arma- 
ture, with  parallel  grouping,  give,  with  a  4  pole  machine,  N  =  40  ;  900 
revolutions  per  minute,  or  15  revolutions  per  second,  with  a  4  pole 
machine,  give  n  =  30  ;  5,000,000  maxwells  are  5  megamaxwells.  Hence, 

E  =  .02  x  30  x  40  x  5  =  120  volts. 

2.  In  a  certain  dynamo  there  are  200  conductors  in  series  in 
each  armature  circuit,  and  the  periodicity  is  5.     The  effective 

85 


86  ELECTKICAL  PROBLEMS 

flux  density  at  the  gap  is  6500  gausses,  and  the  area  of  the  pole 
faces  is  308  square  centimeters  each.  What  is  the  electro- 
motive force? 

3.  Where  4>  is  3.45  megamaxwells,  n  is  6,  and  N  is  128, 
what  is  El 

4.  In  an  8  pole  dynamo  where  the  flux  density  in  the  gap 
is  7000  gausses,  the  area  of  a  pole  face  is  714.3  square  centi- 
meters, the  conductors  in  series  per  armature  circuit  are  400, 
and  the  electro-motive  force  is  500  volts.     What  is  the  speed 
of  rotation? 

5.  If  the  dynamo  in  Problem  4  has  an  armature  resistance 
of  .1  ohm  and  an  output  of  50  kilowatts,  by  what  per  cent  must 
the  flux  density  be  increased  to  maintain  a  terminal   voltage 
of  550? 

6.  A   4  pole  dynamo  has  a  speed  of  500   revolutions  per 
minute,   the   flux   density  is  6.8  kilogausses   over  an  area  per 
pole    of    824    square    centimeters,    and    each    of   its    armature 
circuits  has   168   conductors   in   series.     What  is  the   electro- 
motive force? 

7.  A  railway  generator  is  required  to  give  a  terminal  dif- 
ference of  potential  of  550  volts,  with  a  full  load  output  of 
550  kilowatts.     It  has  10  poles  and  a  speed  of  120  revolutions 
per  minute.     Each  of  its  10  armature  circuits  has  a  resistance 
of  .16   ohm.     What  electro-motive  force  must  be  generated  at 
full  load? 

8.  What  is  the  magnetic  flux  per  pole  in  Problem  7,  with 
110  conductors  in  series  per  armature  circiiit? 

9.  An   8  pole  lighting   generator    built  for  an   output   of 
200   kilowatts   has   76   conductors   in   series,   and  runs   at  200 
revolutions  per  minute.     At  no  load  the  voltage  is  110.    What 
is  the  flux  density  in  the  gap  with  a  polar  area  of  780  square 
centimeters  ? 

10.  The  armature  resistance  of  above  generator  is  .003  ohm. 
With  a  full  load  output  at  125  volts,  what  will  be  the  flux 
density  in  the  gap  ? 


DIRECT   CURRENT   ARMATURES  87 

11.  There  are  8  armature  circuits  in  the  generator  of  Prob- 
lem 10.  With  a  full  load  current  density  of  230  amperes  per  square 
centimeter,  what  is  the  necessary  area  of  an  armature  conductor  ? 

12.  A  6  pole  dynamo  has  an  armature  wound  with  6  circuits, 
each  of  .063  ohm  resistance.     The  flux  per  pole  is  7.8875  mega- 
maxwells,  and  the  speed  300  revolutions  per  minute.     It  is 
required  to  give  a  terminal  electro-motive  force  of  220  volts, 
with  a  full  load  of  150  kilowatts.     What  is   the   number  of 
conductors  in  series  in  an  armature  circuit  ? 

13.  It  is  desired  to  build  a  generator  with  an  electro-motive 
force  of  250  volts  to  run  as  nearly  as  possible  at  500  revolutions 
per  minute ;  the  flux  per  pole  is  2.7  megamaxwells,  the  arma- 
ture has  a  drum  winding  with  2  circuits,  each  of  192  conductors 
in  series.     What  number  of  poles  will  require  the  least  varia- 
tion from  desired  speed,  and  by  what  per  cent  must  the  speed 
be  changed  from  500  to  give  the  exact  voltage? 

14.  A  6  pole  generator  has  a  speed  of  600  revolutions  per 
minute.     What  is  its  periodicity  ? 

15.  The   terminal  electro-motive   force  of   the  generator  of 
Problem  14  is  600  volts,  with  an  output  of  50  kilowatts.     The 
armature  is  wound  as  a  series  drum  of  2  circuits,  each  of  216 
conductors   in  series,   and   of  .2  ohm  resistance.     The  contact 
resistance  of  each  of  the  sets  of  brushes  is  .006  ohm.     What 
electro-motive  force  must  be  generated  in  each  armature  circuit  ? 

16.  The  area  of  a  pole  face  of  the  generator  in  Problem  15 
is  760  square  centimeters.     What  is  the  flux  density  in  the 
air  gap? 

17.  What  size  wire,  B.S.G.,  should  be  used  in  the  armature 
to  give  a  current  density  as  near  315  amperes  per  square  centi- 
meter as  possible  ? 

18.  A    2    pole    direct    current    generator    has    an    induced 
electro-motive  force  of  250  volts,  with  an  output  of  1.15  kilo- 
watts.    There   are   448   conductors  on  the  armature,  and  the 
speed  is  1900  revolutions  per  minute.     What  is  the  magnetic 
flux  through  the  armature  ? 


88  ELECTRICAL  PROBLEMS 

19.  Assuming  a  single  magnetic  circuit  and  a  leakage  coef- 
ficient of  1.35  for  the  dynamo  of  Problem  18,  what  is  the  flux 
density  in  the  yoke  connecting  the  poles  if  it  has  a  rectangular 
area  12  by  16.5  centimeters? 

20.  A   10    pole   generator,   running   at    80    revolutions   per 
minute,  has  a  multiple  drum  winding  of  10  circuits,  each  con- 
taining 128  conductors  in  series.     The  flux  through  each  magnet 
core  is  21.4  megamax wells  at  full  load.     Assuming  a  leakage 
coefficient   of  1.125,  what  is   the  full  load  generated  electro- 
motive force  ? 

21.  In  the  generator  of  Problem  20  the  section  of  an  armature 
conductor  is   .27   by   1.25   centimeters.     What   is   the   current 
density  with  an  output  of  300  kilowatts  at  315  volts? 

22.  If  the  terminal  full  load  electro-motive  force  of  above 
generator  is  315  volts,  what  is  the  resistance  of  the  armature? 

23.  What  is   the   mean  length  of  a   turn  of  the   armature 
winding  if  the  above  resistance  is  at  35°  C.  ? 

24.  The  no  load  electro-motive  force  of  the  above  generator 
is   300   volts.      If  at  this    voltage    the    leakage    coefficient  is 
reduced  to  1.12,   and  the  cross  section  of  the  magnetic  path 
in    the    magnet    frame    is    800    square,    centimeters,    what   is 
the  density? 

25.  In  above  generator  what  is  the  cross  section  of  magnetic 
path  in  the  armature  core,  the  no  load  flux  density  being  11,500 
gausses  ? 

26.  A  6  pole  dynamo  has  the  magnetization  of  its  armature 
reversed  at  the  rate  of  13.75  cycles  per  second.     What  is  the 
speed  ? 

27.  The  generator  in  the  preceding  problem  delivers  a  current 
of  800  amperes  with  an  output  of  10  kilowatts.     The  armature 
resistance  is  .0025  ohm  from  positive  to  negative  brushes.    The 
contact  resistance  of  the  brushes  is  .03  ohm  per  square  inch, 
and  there  are  in  all  30  brushes,  each  with  an  area  of  2  square 
inches.     What    electro-motive    force   must   be   induced   in   the 
armature  ? 


DIRECT  CURRENT  ARMATURES  89 

28.  In  the  generator  above  described  there  are  18  conductors 
in  series  in  each  armature  circuit.     What  is  the  amount  of  the 
magnetic  flux  entering  the  armature  from  any  pole  ? 

29.  If  the  same  generator  have  a  no  load  electro-motive  force 
of  15  volts,  what  is  the  magnetic  flux  in  a  cross  section  of  the 
yoke,  assuming  a  leakage  coefficient  of  1.125? 

30.  The  generator  above  is  multiple  wound  with  6  armature 
circuits.     If  the  current  density  in  the  armature  is  250  amperes 
per   square    centimeter,    what    is    the    area    of    an    armature 
conductor? 

31.  At  full  load  a  motor  absorbs  150  kilowatts  of  electrical 
energy  with  500  volts  at  the  brushes.     The  armature  resistance 
is  .045  ohm.     The  contact  resistance  of  the  brushes  is  .03  ohm 
per  square  inch,  and  there  are  16  brushes  each  1   inch  thick 
and  li  inches  wide.     There  are  4  poles  and  160  conductors  in 
series  per  circuit.     What  is  the  magnetic  flux  per  pole  at  a 
speed  of  450  revolutions  per  minute  ? 


CHAPTER    XVIII 
ALTERNATING   CURRENT   ARMATURES 

When  dealing  with  the  alternating  current  dynamo  the 
fundamental  equation  demands  several  numerical  coefficients 
not  required  in  the  case  of  the  direct  current  dynamo. 

The  first  of  these  is  the  ratio  of  the  virtual  to  the  average 
electro-motive  force  throughout  the  half  period.  With  a  sine 
wave  of  electro-motive  force  this  coefficient  becomes  1.11. 
With  peaked  waves  of  electro-motive  force  it  is  greater  than 
1.11,  while  for  flat  topped  waves  it  is  smaller.  The  value 
1.11  will  be  assumed  in  the  following  problems. 

Another,  called  the  breadth  coefficient  q,  depends  upon  the 
armature  winding,  and  while  in  some  modern  ironclad  alter- 
nators it  is  unity,  in  the  older  surface  wound  armatures  and 
in  modern  ironclad  armatures  with  distributed  windings  its 
value  is  always  somewhat  less  than  unity.  When  other  values 
are  not  specified,  unity  will  be  assumed. 

In  accordance  with  the  above  assumptions,  the  fundamental 
equation  for  alternators  becomes  E  —  2.22  nNq<&~LQ  ~8,  where  <l> 
represents  the  magnetic  flux  per  pole  in  maxwells.  Expressing 
4>  in  megamaxwells,  we  have  E=  .0222  nNq&. 

PROBLEMS 

1.  A   certain  alternator    with    a    periodicity  of    50  has   2.4 
megamaxwells    per  pole  effectively  cut  by  280  conductors  in 
series.     What  is  the  value  of  the  electro-motive  force? 

2.  A    20    pole     alternator    runs    at    240    revolutions    per 
minute  with  636  conductors  in  series  on  the  armature.     The 

00 


ALTERNATING  CURRENT  ARMATURES  91 

area  of  a  pole   face   is   645   square   centimeters   and  the  gap 
density  6.56  kilogausses.     What  is  the  electro-motive  force  ? 

3.  A    12    pole   alternator  has  176    turns   in  series   on  its 
armature    and  is   driven   at  600  revolutions   per   minute.     At 
each  pole  piece  the  air  gap  has  an  area  of  130  square  centi- 
meters and  a  density  of  7500  gausses.     What  electro-motive 
force  is  induced? 

4.  A    3    phase    18    pole    alternator    with    324    conductors 
per  circuit  generates  an  electro-motive  force  of  1330  volts  at 
a  periodicity  of  G2.     What  is  the  magnetic  flux  per  pole  ? 

5.  An  alternator  with  120  poles,  running  at  80  revolutions 
per  minute,    delivers    125    amperes   with    an   output   of    1500 
kilowatts.     With    2880    conductors,    all    in    series,    upon    the 
armature,  -what  is   the  gap  density  with  a  polar  area  of  330 
square  centimeters? 

6.  Keeping  the  flux  per  pole  constant  in  the  generator  of 
Problem  5,  what  electro-motive   force  will  be   induced  if  the 
number  of  poles  is  changed  to  90  and  the  number  of  armature 
conductors  to  2330? 

7.  If  in  the  generator  of  Problem  5  the  number  of  con- 
ductors is  made  2330,  all  else  remaining  the  same,  how  fast 
must    the    machine    run    to    give    an    electro-motive    force    of 
12,000  volts? 

8.  A    6    pole    60    cycle    alternator    with    432    conductors 
upon  the  armature,  all  connected  in  series,  delivers  a  current 
of  5   amperes  with   an  output  of   175   watts.      What  is  the 
speed?     With  a  leakage  coefficient  of  1.15,  what  is  the  flux 
in  the  field  cores  ? 

9.  What    electro-motive   force    is    induced    in    an    8    pole 
alternator  with  1320   conductors  in  series,   an    effective   gap 
density  of   6300   gausses    over   a   polar    area   of    385    square 
centimeters,  and  a  speed  of  750  revolutions  per  minute  ? 

10.  By  what  per  cent  will  the  electro-motive  force  of  a 
dynamo  be  increased  if  both  the  number  of  conductors  and 
the  magnetic  flux  are  increased  50^? 


92  ELECTRICAL  PROBLEMS 

11.  By  what  per  cent  will  the  electro-motive  force  of  the 
generator  of  Problem  9  be  changed  by  increasing  the  speed 
by  90   revolutions  per  minute  and  the  number  of  armature 
conductors  by  180? 

12.  A  10  pole  alternating  current   generator   runs  at    960 
revolutions    per    minute.       Its    armature    is    arranged   for   4 
circuits  of  400  conductors  in  series,  giving  1150  volts,  or  2 
circuits  of  800  conductors  in  series,  giving  2300  volts.     What 
is  the  useful  flux  per  pole? 

13.  The  generator  of  Problem  12  is  rated  at  60  kilowatts. 
If   it   supplies    this    power   on   a   non-reactive    load,   and   the 
resistance  between  brushes   is   1.2   or  4.8   ohms  according  to 
the  method  of  connection,  by  what' per  cent  must  the  flux  be 
increased  to  keep  the  terminal  difference  of  potential  at  1150 
or  2300  volts? 

14.  With   what   size  wire,   B.S.G.,   must   the   generator  of 
Problems  12   and  13  be  wound  to  give  a  copper  area  of  at 
least  785  circular  mils  per  ampere? 

15.  A   16    pole   alternator  running  at  450  revolutions  per 
minute    yields    an   output    of    300    kilowatts    at    110    volts. 
There  are  24  convolutions  in  series  per  circuit.      The  resist- 
ance of  the  armature  and  brush  contacts  being  neglected,  what 
is  the  useful  flux  per  pole? 

16.  A  star   connected   3   phase  alternator  generates  a  cur- 
rent of  250    amperes  in  each  branch  and   has  a  total  output 
of  1000   kilowatts.     If  the   machine   has   36   poles,    384  con- 
ductors per  armature  circuit,  and  an  effective  flux  per  pole  of 
8.2  megamaxwells,  what  is  the  speed? 

17.  What  is  the  output    of  a  single  phase  alternator  sup- 
plying the  same  current  per  phase  at  the  same  terminal  electro- 
motive force  as  that  of  Problem  16  ? 

18.  A   2000    volt   delta  connected  3  phase  alternator  with 
80  poles  runs  at  90  revolutions  per  minute.     The  flux  density 
in  the  air  gap  is  7000  gausses  from  a  pole  face  of  674  square 
centimeters  area.    How  many  conductors  per  circuit  are  needed  ? 


ALTERNATING  CURRENT  ARMATURES  93 

19.    If    the    machine    of    Problem  18  were    wound  as  a    2' 
phase  generator  with   2000  volts  between   the   outside  wires 
of  the  3  wire  line  and  the  speed,  flux,  and  number  of  poles 
kept  the  same,  how  many   conductors  per   circuit   would   be 
needed? 


CHAPTER    XIX 
THE   WINDING  OF  ARMATURES 

The  winding  of  ring  or  Gramme  armatures  is  so  simple  that 
the  student  can  easily  get  a  clear  conception  of  the  subject  from 
the  ordinary  text-book. 

On  the  other  hand,  an  understanding  of  the  far  more  compli- 
cated drum  winding,  especially  in  the  case  of  direct  current 
armatures,  is  hardly  to  be  attained  without  considerable  practice 
in  writing  winding  tables  and  in  drawing  winding  diagrams, 
both  in  end  view  and  in  development. 

Symmetrical  drum  windings  follow  the  law  of  Arnold,  which, 

1  /  C         \ 
symbolically  expressed,  isy  =  -(^-Tif:aj,  where  y  is  the  spacing, 

or  the  distance  expressed  in  elements  of  the  winding,  from  one 
side  of  an  armature  coil  to  the  other,  an  element  of  the  winding 
being  the  group  of  conductors  corresponding  to  one  side  of  an 
armature  coil. 

p  is  the  number  of  pairs  of  poles. 

C  is  the  whole  number  of  conductors  on  the  face  of  the 
armature. 

b  is  the  number  of  conductors  in  an  element  of  the  winding. 

a  is  the  number  of  times  the  current  divides  at  the  negative 
brushes,  or  is  the  number  of  pairs  of  armature  circuits. 

Thus  in  the  case  of  a  direct  current  2  pole  armature  with 
12  coils  of  1  convolution  each,  we  have  p  =  1, 0=  24,  b  =  1,  a  =  1. 


There  are  6  possible  windings,  3  with  each  spacing,  the  tables 
for  2  of  which  are  given  on  the  opposite  page. 


THE  WINDING  OF  ARMATURES  95 

Com.         B  F         Com.  Com.         B  F          Com. 


1 

1 

12 

12 

12 

23 

10 

11 

11 

21 

8 

10 

10 

19 

6 

9 

9 

17 

4 

8 

8 

15 

9 

7 

7 

13 

*24 

6 

6 

11 

22 

5 

5 

9 

20 

4 

4 

7 

18 

3 

3 

5 

16 

2 

2 

3 

14 

1 

1 

1 

14 

2 

o 

3 

16 

3 

3 

5 

18 

4 

4 

7 

20 

5 

5 

9 

22 

6 

6 

11 

24 

7 

7 

13 

2 

8 

8 

15 

4 

9 

9 

17 

6 

10 

10 

19 

8 

11 

11 

21 

10 

12 

12 

23 

12 

1 

Where  the  column  headed  Com.  gives  the  commutator  bars  by 
number,  B  is  the  element  of  the  winding  passing  back  from  a 
commutator  bar  to  the  other  end  of  the  armature  and  F  is  the 
element  of  the  winding  passing  forward  to  the  commutator. 

Thus,  assuming  the  conductors  to  consist  of  24  copper  bars 
already  in  place  on  the  cylindrical  surface  of  the  armature  and 
adopting  a  spacing  of  11,  commutator  bar  No.  1  will  be  con- 
nected to  conductor  No.  1,  the  back  end  of  conductor  No.  1  to 
the  back  end  of  conductor  No.  12,  the  front  end  of  conductor 
No.  12  to  commutator  bar  No.  12,  that  to  the  front  end  of 
conductor  No.  23,  and  so  on,  till  the  front  end  of  conductor 
No.  14  is  connected  to  commutator  bar  No.  1,  thus  completing 
the  connections.  In  like  manner  the  connections  for  a  spacing 
of  13  may  be  traced  out. 

If  the  armature  is  to  be  wound  with  wire  by  hand,  the  coils 
might  be  wound  in  the  following  orders  for  the  two  cases : 

WINDER'S  TABLED 

B  F  B  F  B  F  B  F 

1  12  13  24                     1  14  13  2 

23  10  11  22                     3  16  15  4 

21  8  9  20                     5  18  17  6 

19  6  .    7  18                     7  20  19  8 

17  4  5  16                     9  22  21  10 

15  2  3  14  11  24  23  12 


96 


ELECTRICAL  PROBLEMS 


Thus,  assuming  that  coil  1-12  is  wound  first,  we  then  turn 
the  armature  180°  in  its  forks  and  wind  coil  13—24,  then  coils 
23-10  and  11-22,  and  so  on.  Or  the  winding  may  progress  in 
the  opposite  way  around  the  armature,  thus : 


Com.         B  F  Com. 

1  1  12  2 

2  3  14  3 

3  5  16  4 

Etc. 


Com. 

7 
8 
9 


B 


Com. 


J3        24 

15  2 

17  4 

Etc. 


And  similarly  for  a  spacing  of  13.     After  writing  the  first  two 
lines  of  the  winding  tables  the  rest  can  be  written  directly,  the 


FIG.  20 

same  amount  being  added  each  time  in  going  down  the  vertical 
columns.  The  third  possible  winding  for  each  spacing  would 
be  of  the  wave  type,  to  be  treated  later. 

The  negative  value  of  a,  resulting  in  the  spacing  of  11,  is 
superior  to  the  positive  value,  resulting  in  the  spacing  of  13,  in 
that  it  requires  slightly  less  wire  and  fewer  crossings  at  the  end 
of  the  armature,  thus  lessening  the  danger  of  a  failure  of  insula- 
tion in  the  "  heads." 


THE  WINDING  OF  ABMATUKES 


97 


In  Fig.  20  is  shown  the  end  view  and  in  Fig.  21  the  develop- 
ment of  the  completed  winding  according  to  the  latter  table 
given  for  a  spacing  of  11.  Drawing  in  the  pole  pieces  as  shown 
in  the  end  view  and  development  and  assuming  a  direction  of 
rotation,  we  may  by  Fleming's  rule  determine  the  directions 
of  the  induced  electro-motive  forces  in  the  conductors  that 
are  opposite  the  pole  pieces. 

Now  placing  brushes  upon  the  commutator  bars  that  are 
connected  to  coils  which  lie  near  the  middle  of  the  polar  gaps, 


FIG.  21 


2  and  8  in  Figs.  20  and  21,  we  can  determine  the  polarity  of 
the  brushes,  the  positive  brush  being  taken  as  that  through 
which  the  current  leaves  the  armature. 

Beginning  at  the  negative  brush,  we  may  trace  the  current 
through  the  various  conductors  of  one  of  the  armature  circuits 
to  the  positive  brush,  indicating  the  direction  of  the  current  by 
arrowheads  and  by  dots  and  crosses.  Returning  to  the  nega- 
tive brush,  we  may  trace  and  indicate  the  direction  of  the  current 
through  the  other  armature  circuit  to  the  positive  brush.  If  the 
dynamo  is  multipolar,  the  above  described  process  may  be 
repeated,  starting  in  turn  from  each  negative  brush  and  passing 
in  both  directions  to  an  adjacent  positive  brush. 


98 


ELECTRICAL  PROBLEMS 


PROBLEMS 

1.  Write  the  winding  table  for  and  draw  the  end  view  and 
development  of  a  2  pole  direct  current  drum  armature  with 
36  coils,  each  of  4  convolutions. 

SUGGESTION.  —  There  will  evidently  be  4  conductors  in  an  element  and 
72  elements  in  the  winding.  Instead  of  attempting  to  represent  the 
individual  conductors  in  an  element  as  shown  in  Fig.  22,  it  is  better 
to  represent  the  elements  only  as  shown  in  Fig.  23. 


oo 
oo 

oo 

00 

FIG.  22 

The  formula  for  this  winding  is 
I/  288 


FIG.  23 


1V2X4 


1     =  35  or  37. 


TABLE 


Com. 
1 

36 
35 


B 

1 

71 
69 


F 
36 
34 
32 


Com. 
36 
35 
34 


WINDER'S  TABLE 
B  F         B  F 


1 


36         37 

38        39 

Etc. 


Etc. 


In  drawing  the  end  view  it  is  better  for  the  student  to  make 
the  connections  in  the  order  of  the  winder's  table,  imagining 
as  far  as  possible  that  he  is  actually  putting  wire  upon  the 
armature. 

Any  arrangement  of  conductors  which  results  in  as  many 
armature  circuits  in  multiple  as  there  are  pole  pieces  is  termed 
a  parallel  grouping  of  the  conductors.  Any  arrangement  of 
conductors  which  results  in  but  two  parallel  armature  circuits, 
whatever  the  number  of  poles,  is  called  a  series  grouping.  It  is 
evident  that  in  two  pole  machines  there  is  no  distinction  between 
series  and  parallel  grouping. 


THE  WINDING  OF  ARMATURES  99 

2.  Make  tables  and  drawings  for  a  4  pole  drum  armature 
with  parallel  grouping  of  circuits,  there  being  48  conductors 
and  24  commutator  bars. 

SUGGESTION.  —  In  this  case  there  are  2  pairs  of  poles,  and  since  the 
armature  current  bifurcates  twice,  i.e.,  at  the  2  negative  brushes,  a  is  2. 


The  formula  may  then  be  written  y  =  -(--  T  2  J  =  11 


or  13. 


3.  Make  drawing  for  placing  the  winding  in  Problem  2  upon 
an  armature  with  24  slots. 

SUGGESTION.  —  There  are  half  as  many  slots  as  elements ;  therefore, 
each  slot  will  receive  two  elements.  Assuming  that  the  elements  are 
symmetrically  arranged  in  slots,  1  and  2  in  slot  1,  3  and  4  in  slot  2,  5  and  6 
in  slot  3,  etc.,  the  odd  numbered  elements  being  at  the  bottom  of  the 
slots,  Arnold's  formula  gives  the  correct  winding  for  machine  made  coils ; 
but  in  the  case  of  hand  winding  the  formula  cannot  be  directly  applied. 
If  there  were  only  24  elements,  1  in  each  slot,  the  spacing  would  be 

1  /94         \ 
y  =  -I- 2  J  =  5,  and  the  winder's  table  : 

16  7         12  13         18  19         24 

3         8  Etc. 

Now  with  the  above  tables  and  a  complete  diagram  of  the  armature 
slots,  themselves  numbered  and  containing  the  numbers  representing  the 
elements  of  the  winding,  it  is  easy  to  write  the  winder's  table. 

First  Layer 

1         11  13         23  25         35  37         47 

Etc. 

Second  Layer 

12         22  24         34  36         40  48         10 

Etc. 

The  end  view  drawing  will  be  much  clearer  if  the  inner  and  outer 
layers  are  drawn  in  different  colors ;  or  in  place  of  the  end  view  the  circular 
development  may  be  used  (Fig.  24). 

4.  Write  the  winding  table  for  and  draw  the  end  view  and 
development  of  the  winding  of  a  6  pole  drum  with  parallel 
grouping  of  armature  conductors,  there  being  24  bars  to  the 
commutator. 


100 


ELECTRICAL  PROBLEMS 


5.  Write    the  winding   table    for   a    4  pole    armature  with 
parallel  grouping  of  conductors  and  48  bars  to  the  commutator. 

6.  The  winding  in  Problem  5  is  to  be  placed  by  hand  upon 
an  armature  with  24  teeth.     Write  the  winder's  table. 


FIG.  24 

SUGGESTION.  —  This  will  be  a  case  of  4  layer  \vinding  with  4  elements 
in  each  slot.     Following  the  method  described  under  Problem  3,  we  have 

for  hand  winding : 

First  Layer 

93 


22 


43 


21 


42 


63 


46 


Etc. 


45  49 

Etc. 

Second  Layer 

66  70 

Etc. 

Third  Layer 


69 


90 


7:1. 


94 


18 


THE   WINDING  OF  ARMATURES 


101 


With  machine  wound  coils : 
First  Layer 

1         22 

93         18 

89         14 

Etc. 


Second  Layer 

3         24 
95         20 
91         16 
Etc. 


It  will  be  observed  that  whatever  the  position  of  the  brushes 
upon  the  commutator,  four  successive  coils  always  include  one 
from  each  layer,  thus  insuring  an  equality  of  resistance  in  the 
four  armature  circuits  and  a  constancy  of  resistance  at  all  times. 


3  |  4  |  5  I  6  I  7  |  8  |  9  |  10  |  11  |  12  |  13  |  14  |  15  |  16  |  17  |  18  |  19  |  20  |  21 1  22  |  1  |  2  | 


FIG.  25 

It  is  of  course  evident  that  in  any  armature  circuit  the 
spacing  must  be  such  that  two  successive  elements  of  the 
winding  will  be  similarly  situated  with  respect  to  adjacent,  and 
therefore  opposite,  field  poles.  In  the  windings  hitherto  consid- 
ered the  connector  on  the  front  or  commutator  end  of  a  coil  has 
lapped  back  from  the  F  element  of  a  given  coil  to  the  B  end  of 
the  coil  next  in  order,  before  or  behind,  around  the  surface  of 
the  armature.  This  arrangement  is  known  as  lap  winding,  and 
is  clearly  shown  in  the  development  diagram  (Fig.  21). 

It  is  evident,  however,  that  the  condition  that  the  electro- 
motive forces  in  the  various  elements  of  an  armature  circuit 


102  ELECTRICAL  PROBLEMS 

shall  be  cumulative  will  be  equally  well  satisfied  by  connecting 
the  F  element  of  a  given  coil  to  the  B  element  of  a  coil  situated 
about  2  poles  in  advance  of  the  first  coil.  This  arrangement 
is  known  as  wave  winding,  and  is  shown  in  the  development 
diagram  (Fig.  25),  which  is  the  development  of  the  winding  of 
Problem  7. 

7.  Write  the  winding  table  for  and  draw  the  end  view  and 
development  of  a  6  pole  wave  wound  armature  having  44  con- 
ductors arranged  in  series  grouping. 

SUGGESTION.  —  With  series  grouping  there  can  be  but  2  armature  cir- 
cuits ;  hence  a  =  \ .     The  formula  for  spacing  becomes  //  =  -  (  —  :p  1  j  =  7. 

8.  Write  the  table  for  and  draw  the  development  of  a  4 
pole  wave  winding  having  54  conductors,  arranged  with  series 
grouping,  and  connected  to  a  27  part  commutator. 

9.  Write  winding  tables  for  a  6  pole  wave  wound  armature 
with  88  elements  connected  in  2  parallel  circuits. 

10.  In  a  6  pole  wave  wound  armature  with  elements  connected 
in  series  grouping  the  spacing  is  27.     How  many  elements  are 
there  in  the  winding  ? 

11.  Write  the  tables  for  a  4  pole  armature  with  series  group- 
ing, there  being  92  conductors  arranged  in  23  coils,  and  wound 
in  2  layers  in  23  slots.     Draw  enough  slots  in  end  view  to 
show  arrangement  of  elements  in  the  slots. 

12.  Write  tables  for  and  draw  end  view  and  development  of 
a  6  pole  wave  wound  armature  with  series  grouping  of  circuits, 
there  being  124  elements  arranged  4  in  a  slot.    Draw  teeth  and 
number  elements  in  end  view. 

Describe  the  following  windings  : 


Solution.     An  inspection  of  the  formula  shows  that  this  is  an  8  pole 
•winding  with  90  conductors  in  45  coils  in  series  grouping. 


THE  WINDING  OF  ARMATURES  103 

JL  /   J.  />    i  •+  n  -^  /    j.  **  v/  o\ 

—  O     I . 


I/  616 


X 

Cases  of  mixed  lap  and  wave  windings  have  been  designed  to 
meet  special  conditions.  In  general  they  have  no  particular 
advantage  over  simple  lap  or  wave  windings. 

19.  Below  is  given  the  table  for  a  4  pole  mixed  lap  and 
wave  winding  with  series  grouping,  designed  for  a  particular 
purpose.  There  are  84  conductors  arranged  in  21  slots.  Draw 
the  end  view  and  development,  place  the  poles,  trace  out  the 
currents,  and  locate  the  brushes. 


Com. 

B 

F 

Com. 

B 

F 

Com. 

B 

F 

Com. 

B 

F  Com. 

I 

1 

22 

12 

43 

64 

23 

45 

66 

34 

3 

24 

3 

3 

5 

26 

14 

47 

68 

25 

49 

70 

36 

7 

28 

5 

5 

9 

30 

16 

51 

72 

27 

53 

74 

38 

11 

32 

7 

7 

13 

34 

18 

55 

76 

29 

57 

78 

40 

15 

36 

9 

9 

17 

38 

20 

59 

80 

31 

61 

82 

42 

19 

40 

11 

11 

21 

42 

22 

63 

84 

33 

65 

2 

2 

23 

44 

13 

13 

25 

46 

24 

67 

4 

35 

69 

6 

4 

27 

48 

15 

15 

29 

50 

26 

71 

8 

37 

73 

10 

6 

31 

52 

17 

17 

33 

54 

28 

75 

12 

39 

77 

14 

8 

35 

56 

19 

19 

37 

58 

30 

79 

16 

41 

81 

18 

10 

39 

60 

21 

21 

41 

62 

32 

83 

20 

1 

The  winding  of  an  alternating  current  armature  is  very  easily 
represented ;  care  should  be  taken,  however,  that  the  electro- 
motive forces  in  the  various  coils  of  the  same  armature  circuit 
shall  all  be  cumulative. 


104 


ELECTRICAL  PROBLEMS 


Fig.  26  represents  the  development  of  a  4  pole  single  phase 
winding  with  32  conductors  in  a  single  circuit. 

Fig.  27  represents  the  development  of  a  4  pole  2  phase  wind- 
ing with  24  conductors  per  phase,  the  2  circuits  being  entirely 
independent. 

Fig.  28  represents  the  development  of  a  4  pole  3  phase  wind- 
ing with  24  conductors  per  phase,  arranged  in  star  grouping  of 
the  phases. 

20.  Draw  the  development  and  end  view  of  the  winding  for 
a  6  pole  single  phase  armature  with  36  conductors  in  1  circuit. 


FIG.  20 


21.  Draw  the  development  and  end  view  of  the  winding  for 
an  8  pole  2  phase  armature  with  32  conductors  per  phase. 

22.  Draw  the  development  and  end  view  of  the  winding  of 
a  6  pole  3  phase  armature  with  24  conductors  per  phase,  the 
phases  being  connected  in  star  grouping. 

23.  Connect  the  circuits  of  Problem  23  in  mesh  grouping. 

SUGGESTION.  —  Number  the  circuits  1,  2,  3.  Let  poles  be  drawn  in  such 
position  that  circuits  1  and  2  at  least  are  generating  electro-motive  forces ; 
now  connect  the  negative  terminals  of  circuits  1  and  2,  the  positive  termi- 
nals of  circuits  1  and  3,  and  the  positive  terminal  of  2  to  the  negative 
terminal  of  3.  These  3  junctions  should  be  connected  to  the  3  collecting 
rings. 


THE   WINDING  OF  AllMATUEES 


105 


L L 


FIG.  27 


FIG.  28 


106  ELECTRICAL  PROBLEMS 

24.  Draw  the  development  and  end  view  of  a  16  pole  3  phase 
winding  with  star  grouping  of  the  circuits  and  576  conductors 
in  all. 

SUGGESTION.  —  As  in  the  case  of  a  direct  current  armature  the  various 
conductors  forming  one  side  of  a  coil  may  be  conceived  of  as  forming  an 
element  of  the  winding,  and  only  these  elements  need  be  represented  in  the 
drawing. 

25.  Draw  the  development  of  a  3  phase  12  pole  winding  with 
mesh  grouping  of   the   armature  circuits  and  144   conductors 
per  phase. 

26.  Draw  the  end  view  of  the  winding  of  a  3  phase  8  pole 
armature  with  48  slots  and  star  grouping  of  the  circuits. 

27.  Draw  the  development  of  the  winding  of  a  3  phase  4  pole 
armature  with  8  coils  per  phase,  the  circuits  being  connected  in 
mesh  grouping. 


CHAPTER    XX 
ARMATURE   REACTIONS 

If  a  generator  is  to  maintain  its  voltage  at  full  load,  it  must 
of  course  generate  additional  volts  enough  to  force  the  full  load 
current  through  the  impedance  of  the  armature  conductors  and 
brush  contacts.  If,  at  the  same  time,  the  brushes  of  a  direct 
current  generator  are  given  a  forward  lead  to  prevent  sparking, 
the  demagnetizing  action  of  the  armature  currents  within  the 
angle  of  lead  weakens  the  field  and  increases  the  leakage 
coefficient.  Since  the  magnetic  flux  may  be  assumed  to  enter 
and  leave  the  armature  almost  entirely  by  way  of  its  face,  it  is 
evident  that  the  magnetizing  effect  of  the  conductors  depends 
only  on  the  amount  and  distribution  of  the  currents  in  them 
and  not  at  all  on  the  method  of  connecting  them  at  the  ends  of 
the  armature.  For  our  present  purpose  we  may  therefore  assume 
the  conductors  to  be  connected  as  shown  in  Fig.  29,  where  it  is 
seen  that  the  demagnetizing  ampere  turns  per  pole  are  equal  to 
the  product  of  the  number  of  conductors  contained  in  the  angle 
of  lead  by  the  current  in  each  conductor.  The  currents  in  the 
other  conductors  more  or  less  distort  the  field  but  do  not 
materially  alter  the  total  flux.  Hence  the  field  ampere  turns 
required  to  force  the  larger  flux  needed  for  the  higher  electro- 
motive force  through  the  reluctance  of  the  magnetic  circuit  must 
be  recalculated,  and  to  the  result  thus  obtained  must  be  added 
the  back  ampere  turns  of  the  armature,  the  whole  being  affected 
by  the  new  value  of  the  leakage  coefficient.  The  above  state- 
ments apply  equally  to  a  direct  current  motor  with  a  backward 
lead  to  the  brushes. 

107 


108 


ELECTRICAL  PROBLEMS 


PROBLEMS 

1.    A  10   pole   direct  current  armature  has   180   slots  with 
4  conductors  in  each.    What  are  the  demagnetizing  ampere  turns 


i  Brushes  at  Neutral  Position. 
•  Brushes  at  Full  LoaoLPosition. 
FIG.  29 


if   each   conductor   carries   200   amperes   and  the   brushes   are 
advanced  T\  of  the  polar  pitch? 

2.    A   2   pole   direct  current  armature  has   88   convolutions 
wound  upon  it.     What  are   the  demagnetizing   ampere   turns 


ARMATURE   REACTIONS 


101) 


when    the    brushes    are    advanced    10°   with    a    load    of    100 
amperes  ? 

3.    A  4  pole  generator  with  4  armature   circuits   is  wound 
with  640  conductors  in  40  slots.      What  are  the  demagnetizing 


FIG.  30 


ampere  turns  with  an  output  of  120  amperes,  the  brushes  being 
advanced  through  an  arc  of  9°  ? 

4.  A  12  pole,  1500  kilowatt,  600  volt  railway  generator  is 
wound  with  1392  armature  conductors  in  multiple  grouping. 
What  lead  of  brushes  in  degrees  of  arc  will  produce  2600 
demagnetizing  ampere  turns  per  pole  at  full  load  current? 


110  ELECTRICAL   PROBLEMS 

5.  A  12.5  kilowatt  input,  4  pole,  250  volt  motor  with  series 
wound  armature  has  560  conductors  in  35  slots.     At  full  load 
the  brushes  are  given   a  backward   lead  of    7°.     How  many 
demagnetizing  ampere  turns  are  thereby  produced  at  each  pole  ? 

If  the  current  in  an  alternating  current  generator  lags  in 
phase  behind  the  electro-motive  force,  then,  as  shown  in  Fig.  30, 
the  armature  current  will  set  up  a  magneto-motive  force  oppos- 
ing that  of  the  field  coils,  because  each  conductor  has  turned 
beyond  the  middle  of  the  polar  arc,  the  position  of  maximum 
electro-motive  force,  before  carrying  its  maximum  current,  and 
accordingly  the  resultant  flux  is  diminished.  Evidently  a  lead- 
ing current  will  increase  the  resultant  magneto-motive  force. 
The  reactions  of  a  synchronous  motor  are  opposite  to  those  of 
a  generator. 

The  amount  of  this  reaction  for  a  given  angle  of  lag  will 
depend  somewhat  upon  the  distribution  of  the  armature  conduc- 
tors and  upon  the  shape  of  the  field  poles,  but  it  may  be  assumed 
to  vary  approximately  with  the  sine  of  the  angle  of  lag,  the  net 
effect  being  the  number  of  armature  ampere  turns  per  pole 
multiplied  by  the  sine  of  this  angle. 

In  alternating  current  armatures  also  we  have  not  only  the 
resistance  drop  but  a  reactive  drop  due  to  the  inductance  of  the 
armature  winding.  The  vector  sum  of  these  two,  which  are  of 
course  in  quadrature,  is  the  impedance  drop  and  is  sometimes 
quite  large. 

6.  A  star  connected,  3  phase,  64  pole  generator  running  at 
50  cycles  per  second  and  supplying  1000  kilowatts  at  5000  volts 
has  the  armature  wound  with  32  coils  per  phase,  each  of  12  con- 
volutions.    At  full  load  with  unit  power  factor,  the  total  drop 
was   observed  to  be   8%   and  the  measured  resistance  of   the 
armature  was  .34  ohm  per  phase.      Determine  the  value   per 
phase  of  the  drop  due  to  resistance,  that  due  to  inductance,  and 
the  inductance  of  the  winding. 

7.  If  the  armature  in  Problem  6  is  short  circuited,  by  what 
angle   will   the   current  lag  behind   the   electro-motive   force? 


AftMATUKE  REACTIONS  111 

Consider  the  self-induction  of  the  armature  to  be  the  same  as 
at  full  load. 

8.  In  Problem  7  what  will  be  the  demagnetizing  ampere 
turns   per  pole   with   full   load   current   flowing   in    the    short 
circuited  armature? 

9.  If  the  above  generator  supplies  full  load  current  with  a 
lag  of  30°,  how  many  demagnetizing  ampere  turns  psr  pole  will 
it  produce? 

10.  A  14  pole  single  phase  generator  runs  with  a  power  fac- 
tor of  .7  and  an  apparent  output  of  180  kilowatts  at  30,000 
volts.  The  armature  is  wound  with  14  coils  of  156  convolutions 
each.  How  many  demagnetizing  ampere  turns  are  thereby 
produced  ? 

XOTE.  —  By  apparent  output  is  meant  the  product  of  volts  and  amperes. 


CHAPTER   XXI 
FIELD   WINDING 

The  analogue  for  the  magnetic  circuit  of  the  law  of  Ohm  is 

magneto-motive  force 

the   magnetic  flux  <!>,=—        —. =—          —  •    The  magnetic 

magnetic  reluctance 

force  is  expressed  by  4  TT  times  the  absolute  current  turns,  or  by 
y^  times  the  ampere  turns.     The  magnetic  reluctance  may  be 

reluctivity  X  length  of  path 

expressed  by  -  — ,  or  more  usually  by 

area 

length  of  path  4  TT  a.  t.  .8  <&l 

;  hence  <£  = = —  ,  or  a.  t.  =  -      — ,  where 


-I       .-•    •    .  J        *>*.\J*.*.\S\S        -^T       7  7     V^J.         «_*•        I/*       A 

permeability  X  area  I  AJJL 

10 — 7 
fjiA 

<£) 
a.  t.  signifies  ampere   turns.     The   flux  density  —  is  usually 

jrL 

denoted  by  B,  so  that  for  any  section  of  a  magnetic  path  we 
have   a.  t.  =  .8  B  —  =  .8  HI,  where   a.  t.  are   the  ampere  turns 

needed  to  produce  the  flux  of  density  B  gausses  through  a 
section  of  length  I  and,  at  that  density,  permeability  p. 

The  permeability  of  air  and  all  materials  other  than  iron  and 
steel  used  in  the  construction  of  dynamos  is  practically  unity ; 
accordingly  in  them  we  have  a.  t.  =  .8  Bl. 


PROBLEMS 


1.  How  many  ampere  turns  are  required  to  establish  a  flux 
of  2.5  megamaxwells  across  an  air  gap  .5  centimeter  long  and 
with  an  area  of  412  square  centimeters? 

112 


FIELD  WINDING 

Density  Gausses 


113 


114  ELECTRICAL  PROBLEMS 

2.  The  average  length  of  the  flux  path  through  an  armature 
core  is  28  centimeters,  and  its  cross  section  is  100  square  centi- 
meters.    Selecting  a  value  of  H  from  the  curve  for  stampings 
(Fig.  31)  corresponding  to  the  value  of  B  for  the  flux  of  Prob- 
lem 1,  and  remembering  that  only  half  the  given  length  of  path 
is  to  be  considered  as  belonging  to  one  pole,  and  that  half  the 
flux  from  a  pole  goes  in  each  direction  through  the  armature, 
calculate  the  ampere  turns  per  pole  required  for  the  armature 
core. 

3.  There  are  9  teeth  under  each  pole,  each  with  an  area  of 
14  square  centimeters,  and  a  radial  depth  of  2.75  centimeters. 
Determine  the  ampere  turns  per  pole  needed  to  establish  the 
above  flux  of  2.5  megamaxwells  in  the  teeth. 

4.  The  magnet  core  of  cast  steel  has  a  length  of  16.5  centi- 
meters and  a  section  of  250  square  centimeters.     Assuming  a 
leakage  coefficient  of  1.125,  how  many  ampere  turns  per  pole 
are  needed  to  set  up  the  required  flux  in  the  magnet  cores  ? 

5.  The  length  of  the  flux  path  in  the  cast-iron  yoke  ring  is 
57  centimeters,  and  its  section  is  322  square  centimeters.    Bear- 
ing in  mind  that  the  leakage  coefficient  applies  to  the  yoke  as 
well  as  to  the  magnet  core,  and  that  the  flux  divides  in  the 
yoke  as  in  the  armature  core ;   also  that  each  field  coil  forces 
the  flux  through  only  half  the  yoke,  calculate  the  ampere  turns 
per  pole  required  for  the  yoke. 

6.  Summing  up  the  results  of  the  five  preceding  problems, 
what  number  of  ampere  turns  is  required  for  each  field  coil? 
If  the  dynamo  has  4  poles,  how  many  ampere  turns  are  required 
for  the  whole  machine  ? 

Having  found  the  whole  number  of  ampere  turns,  we  may 

pL 
determine  the  proper  size  of  wire  as  follows:  E=-—->  where 

R  is  the  total  resistance  of  the  field,  L  the  total  length  of  the 
winding,  equal  to  I  times  the  number  of  turns,  I  being  the 
length  of  a  turn,  p  the  resistance  of  a  centimeter  of  wire  with 
an  area  of  1  square  millimeter,  at  the  temperature  which  the 


FIELD  WINDING  115 

field  reaches,  and  A  the  area  of  the  wire  in  square  millimeters. 

pL      pi  X  no.  turns .    Ipl    x  no.  turns      pi  X  a.  t. 
Then    A—g*  -g-  -^-  -^— , 

where  ^  is  the  voltage  exciting  the  field  and  /  the  field  current. 
At  0°  C.  p  =  159  x  ID"0.  The  temperature  of  the  coils  is  likely 

to  reach  60°  C.,  at  which  p  -197  x  lQ-°  and  A  =  -—-^- 

10  IL 

Having  found  A,  we  find  from  a  wire  table  (page  142)  the  size 
of  wire  to  use.  If  it  comes  between  two  standard  sizes,  in 
default  of  having  a  special  wire  drawn,  the  larger  of  the  two 
must  be  used.  Enough  turns  must  be  supplied  so  that  the 
current  density  will  not  exceed  the  limits  of  100  to  160  amperes 
per  square  centimeter. 

In  the  United  States  calculations  are  frequently  made  in 
English  dimensions,  into  which  the  formula  can  be  changed  by 
a  change  of  the  numerical  constant. 

7.  Assuming  that  14,940  ampere  turns  are  needed  for  the 
whole  field  of  above  generator,  and  that  we  are  limited  to  the 
standard  sizes  of  wire,  what  size  will  be  required  for  the  field 
if  the  mean  length  of  a  turn  is  78  centimeters  and  the  electro- 
motive force  is  600  volts? 

8.  If  the  winding  space  for  the  field  coils  is  13.5  centi- 
meters long  on  each  pole  piece  and  the  diameter  of  the  insu- 
lated wire  is  .95  millimeter,  how  deep  must  the  winding  be  that 
the  current  density  may  not  exceed  130  amperes  per  square 
centimeter  ? 

9.  The  field  coils  being  joined  in  series,  what  will  be  their 
resistance  at  40°  C.  ?    at  60°  C.  ? 

10.  What  resistance  must  be  in  the  field  rheostat  at  60°  C. 
and  at  40°  C.,  in  order  that  in  each  case  there  may  be  14,940 
ampere  turns  required  for  the  no  load  electro-motive  force  of 
600  volts  to  be  generated  by  the  armature? 

11.  What  is  the  I*R  loss  in  the  field  at  40°  C.  ?  at  60°  C.  ? 
Field  coils  require,  on  the  average,  for  each  watt  radiated 

400    square    centimeters    divided   by   the    number   of   degrees 


116  ELECTRICAL   PROBLEMS 

Centigrade   difference  in   temperature  between   the  surface  of 
the  coil  and  the  surrounding  air. 

12.  If  at  no  load  the  temperature  of  the  field  coils  is  40°  C. 
and  the  machine  is  generating  full  load  volts,  what  will  be  the 
rise  in  temperature  of  the  field  coils  above  the  surrounding  air? 

13.  In  the  case  of  the  above  generator  the  total  armature  and 
brush  resistance  is  2  ohms,  the  leakage  coefficient  at  full  load  is 
1.14,  and  the  armature  current  is  15  amperes.     Calculate  the 
ampere  turns  needed  to  drive   the  full  load  flux  through  the 
magnetic  circuit. 

14.  There  are  2  armature  circuits  of  710  turns  each,  and  the 
lead  of  the  brushes  with  a  load  of  15  amperes  is  10°.    Calculate 
the  demagnetizing  ampere  turns  per  pole. 

15.  What  will  be  the  total  ampere  turns  per  pole  required  at 
full  load? 

16.  Recalculate  the  size  of  wire  needed  in  the  field. 

17.  Assuming  the  resistance  of  the  field  coils  to  be  that  at 
55°  C.,  calculate  the  full  load  temperature  rise  above  surround- 
ing air  with  the  current  density  allowed  at  no  load. 

18.  If  the  generator  above  treated  is  compound  wound,  the 
shunt  coils  supplying  only  the  no  load  ampere  turns,  what  will 
be  the  area  and  number  of  turns  of  the  series  winding  for  con- 
stant potential  working,  not  .considering  the  resistance  of  the 
series  coils?    Allow  a  density  in  the  series  coils  of  180  amperes 
per  square  centimeter,  and  assume  that  f  of  the  current  flows 
through  the  series  coils. 

19.  A  3  phase,  mesh  wound,   760  kilowatt  alternator  with 
64  poles  runs  at  79.7  revolutions  per  minute.     The  terminal 
voltage   is   2200.     The   pole  pieces  are   25   centimeters   long, 
parallel  to  the  shaft,  with  an  arc  15  centimeters  long,  and  the 
air  gap  is  9  millimeters  long.     The  magnet  cores  are  of  cast 
steel,  14  centimeters  long,  with  an  area  of  200  square  centi- 
meters,   bolted   on   a  cast-iron    field    ring   whose  external  and 
internal  diameters  are  565  and  552  centimeters,  respectively, 
and  whose  length  is  32  centimeters,  parallel  to  the  shaft.    The 


FIELD  WINDING  117 

armature  is  wound  with  32  coils  per  phase,  each  of  12  convolu- 
tions. On  account  of  low  density  we  may  neglect  the  reluc- 
tance of  the  magnetic  path  in 'the  armature  core  and  teeth. 
Find  how  many  ampere  turns  are  needed  at  no  load,  with  a 
leakage  coefficient  of  1.1. 

20.  The  field  -winding  consists  of  50  turns  per  pole  of  copper 
ribbon.     What  is  the  no  load  exciting  current  ? 

21.  At  full  load  with  unit  power  factor  the  drop  is  2.81J&. 
What  is  the  armature  resistance  per  phase,  hot? 

22.  At  full  load  current  with  a  non-reactive  external  circuit 
the  drop  is  4/o.     Find  the  inductance  of  the  armature. 

23.  Calculate    the    exciting    current    needed   for   full    load 
current  with  the  armature  short  circuited. 

24.  Calculate  the  full  load  exciting  current  on  a  non-reactive 
circuit. 

25.  With  a  power  factor  of  .85,  what  exciting  current  will  be 
required  to  maintain  the  terminal  voltage  of  2200  with  full 
load  current  and  inductive  load? 


CHAPTER    XXII 
THE    TRANSFORMER 

Fig.  32  represents  the  diagram  of  a  constant  potential  trans- 
former with  secondary  current  in  phase  with  the  secondary 
electro-motive  force,  the  transformation  ratio  being  assumed 
unity.  0<&  represents  the  flux,  produced  by  the  exciting  cur- 
rent On,  composed  of  magnetizing  component  Om  and  hysteretic 
component  mn  in  quadrature.  The  alternating  flux  0<&  induces 
the  electro-motive  forces  e\  and  ev  respectively,  in  primary  and 
secondary  coils.  The  secondary  electro-motive  force  e2  sets  up 


FIG.  32 


the  current  J2,  requiring  OL2  volts  to  overcome  the  impedance  of 
the  secondary  winding ;  there  results  the  difference  of  potential 
E2  at  secondary  terminals,  the  vector  difference  between  Oe2 
and  OL2.  To  neutralize  the  magnetizing  action  of  0/2  amperes 
in  the  secondary,  there  is  required  nl^  amperes  in  the  primary, 
which  combined  with  the  exciting  current  On  produces  the  total 
primary  current  OIr  To  overcome  the  impedance  of  the  pri- 
mary, OLl  volts  are  required.  To  overcome  Oe\  volts  induced  in 

118 


THE  TRANSFORMER  119 

the  primary,  Oe^  volts  are  required ;  hence  there  must  be  OE^ 
volts  impressed  upon  the  primary. 

The  fundamental  equation  of  the  transformer  is 

E=  ^27rnN<&  X  10~8, 

where  E  is  the  electro-motive  force  induced  in  N  convolutions 
about  an  iron  core  whose  total  flux  <I>  is  alternating  with  a 
periodicity  n. 

Since  in  transformers  with  closed  iron  magnetic  circuits  the 
current  and  electro-motive  force  are  practically  in  phase  except 
when  the  load  is  very  light,  we  may  assume  that  El  for  the 
primary  coil  is  equal  to  the  impressed  electro-motive  force  minus 
the  primary  ohmic  drop,  and  that  Ez  for  the  secondary  coil  equals 
the  difference  of  potential  at  the  secondary  terminals  plus  the 
secondary  ohmic  drop. 

<I>  may  of  course  be  expressed  as  B  max  x  -4,  A  being  the  area 
of  the  magnetic  circuit  and  Bmax  the  maximum  value  of  the 
induction  density  during  the  cycle.  The  fundamental  equation 
then  becomes 

Ex  108 


E  =  V'2  TrnNAB  max  X  10  ~8,  or  AN  = 


4.44  n  B  max 


Under  practical  conditions  E  and  n  are  fixed  and  B  max  should 
be  such  that  with  the  periodicity  employed  the  loss  by  hys- 
teresis and  eddy  currents  will  not  much  exceed  .015  watt  per 
cubic  centimeter. 

The  approximate  flux   densities    commonly  employed  with 
different  periodicities  are  given  in  the  following  table : 

Frequencies  B  max 

125 3000  gausses 

100 3500 

80 4000 

60 5000  to  6000 

50 6000  to  7000         « 

25  10,000  to  12,000        « 


120  ELECTRICAL  PROBLEMS 

An  inspection  of  the  fundamental  equation  shows  that, 
neglecting  ohmic  drop  and  magnetic  leakage,  the  ratio  of  the 
primary  to  the  secondary  voltage,  or  the  ratio  of  transformation, 
is  the  ratio  of  the  primary  to  the  secondary  convolutions. 

PROBLEMS 

1.  A  transformer  for  supplying  lamps  at  110  volts  is  con- 
nected   to   primaries    at    2200    volts.     What    is    the    ratio    of 
transformation  ? 

Solution.     Neglecting  ohmic  drop  and  magnetic  leakage,  we  have 

^      2200 
E2        110 

2.  If  the  above  transformer  has   1200  convolutions  in  the 
primary,  how  many  are  in  the  secondary? 

3.  A  step  up  transformer  has  200  convolutions  in  the  primary 
and  a  transformation  ratio  of  .1.     Assuming  1100  volts  to  be 
impressed  upon  the  primary,  what  is   the  secondary  electro- 
motive force  and  what  the  number  of  secondary  convolutions  ? 

4.  A  system  consists  of  a  generator,  a  step  up  transformer 
having  300  primary  and  4500  secondary  convolutions,  a  trans- 
mission line,  and  a  step  down  transformer  having  5625  primary 
and  75  secondary  convolutions.     The  step  down  transformer  is 
to  supply  current  at  390  volts.    Neglecting  all  losses,  what  must 
be  the  electro-motive  force  of  the  generator? 

5.  In  the  electrical  system  of  Problem  4  assume  that  there  is 
at  full  load  a  2%  loss  of  pressure  in  each  transformer  and  a  10% 
drop  on  the  line.    By  what  per  cent  must  the  generator  electro- 
motive force  be  increased  to  keep  the  terminal  electro-motive 
force  of  the  secondary  at  390  volts  ? 

6.  A  step  down  transformer  with  a  transformation  ratio  of 
10   supplies  current  at  200  volts  to  a  single  phase  motor  of 
3  ohms  equivalent  resistance  and  4  ohms  equivalent  reactance. 
The  primary  resistance  of  the  transformer  is  2  ohms  and  the 


THE  TRANSFORMER  121 

primary  reactance  5  ohms;  the  secondary  resistance  is  .02  ohm 
and  the  secondary  reactance  .05  ohm.  The  primary  exciting 
current  is  .2  ampere  and  the  core  loss  250  watts.  By  means  of 
the  transformer  diagram  determine  the  magnitude  and  phase 
relations  of  the  primary  electro-motive  force  and  current. 

7.  Determine  the  magnitude  and  phase  relations  of  the  pri- 
mary electro-motive  force  and  current  when  the  above  transformer 
is  furnishing  current  for  an  overexcited  synchronous  motor  of 
4  ohms  equivalent  resistance  and  3  ohms  equivalent  reactance. 

8.  Determine  the  magnitude  and  phase  relations  of  the  pri- 
mary electro-motive  force  and  current  when  the  above  trans- 
former   is    furnishing    current    for    200    16    candle    3^    watt 
incandescent  lamps. 

9.  A  60  cycle  15  kilowatt  transformer  for  a  2080  volt  circuit 
has  1360  convolutions  in  the  primary  and  a  transformation  ratio 
of  20.     The  mean  length  of  the  magnetic  circuit  is  179  centi- 
meters and  its  section  102  square  centimeters.      Assuming  a 
permeability  of  2600,  a  hysteretic   constant  of  .0015  erg  per 
gauss  per  cubic  centimeter  per  cycle,  and  that  the  eddy  current 
loss  is  a  quarter  of  the  hysteretic  loss,  what  is  the  exciting  cur- 
rent and  what  is  the  core  loss  ? 

10.  What  is  the  no  load  power  factor  of  the  transformer  of 
Problem  9  ? 

11.  The  primary  winding  is  of  No.  9  wire,  B.S.G.,  the  mean 
length  of  a  convolution  being  59  centimeters.     The  secondary 
has  50  centimeters  per  turn  and  is  built  up  of  flat  strips  having 
an  aggregate  section  of  1.32  square  centimeters.    At  a  tempera- 
ture of  30°C.,  what  is  the  primary  and  what  the  secondary 
resistance  ? 

12.  What  is  the  full  load  current  density  in  each  winding  of 
the  transformer  in  Problem  11  ? 

13.  What  is  the  IR  drop  in  each  of  the  above  coils  at  full 
load? 

14.  What  is  the  total  full  load  copper  loss  in  the  above  trans- 
former ? 


122  ELECTRICAL  PROBLEMS 

15.  Calculate  the  efficiency  of  this  transformer  at  each  quarter 
of  full  load. 

16.  A  125  cycle,  5  kilowatt,  1100  volt  transformer  supplies 
current  to  a  110  volt  lighting  circuit.     The  primary  consists  of 
1060  convolutions  of  No.  12  wire,  B.S.G.    The  current  density 
in  the  secondary  at  full  load  is  .8  that  in  the  primary.     The 
mean  length  of  the  magnetic  circuit  is  98.5  centimeters  and  its 
effective  area  69  square  centimeters.     The  mean  length  of  a 
primary  convolution  is  42  centimeters  and  of  a  secondary  con- 
volution  36   centimeters.       Determine   the   efficiency  at    each 
quarter  of  full  load,  at  a  temperature  of  35°  C. 

17.  Assuming  a  constant  secondary  terminal  voltage  of  110, 
determine  the  primary  voltage  at  each  quarter  of  full  load. 

18.  A  25  cycle,  2200  kilowatt  transformer  to  transform  from 
22,000  to  2200  volts  has  a  double  magnetic  circuit  built  up  of 
plates  112  by  57  centimeters,  with  a  rectangular  hole  80  by  25 
centimeters  and  a  triangular  piece  7  centimeters  on  each  of  the 
shorter  sides  cut  from  each  corner.     On  each  leg  of  the  copper 
coils  enough  of  these  plates  are  assembled  to  make  an  aggregate 
thickness  of  iron  of  100  centimeters.     There  are  600  convolu- 
tions in  the  primary  coil.     What  is  the  maximum  value  of  the 
flux  density? 

19.  Assuming  a  hysteretic' coefficient  of  .00152  and  an  eddy 
current  loss  half  as  large  as  the  hysteresis  loss,  what  is  the  total 
iron  loss  at  no  load  ? 

20.  Assuming  the  length  of  a  convolution  in  both  primary 
and  secondary  to  be  360  centimeters,  what  will  be  the  full  load 
copper  loss  with  a  current  density  of  225  amperes  per  square 
centimeter,  the  temperature  being  45°  C.  ? 

NOTE.  —  The  transformer  is  artificially  cooled,  and  hence   this  high 
current  density  is  allowable. 

21.  What  will  be  the  efficiency  of  the  transformer  of  Prob- 
lem 20  at  each  quarter  of  full  load? 

22.  To   a   400   volt   Thompson   compensator   are  connected 
40   100   volt   incandescent   lamps,   each    requiring   1    ampere. 


THE  TRANSFORMER  123 

There  are  20  lamps  in  the  first  branch,  15  in  the  second,  and  5 
in  the  third.  Neglecting  all  losses,  determine  and  draw  diagram 
showing  the  current  in  each  coil  of  the  compensator  and  its 
direction  as  compared  with  the  impressed  electro-motive  force. 

23.  Assuming  that  the  core  loss  in  above   compensator  is 
80  watts  and  the  total  resistance  of  the  winding  is  1  ohm,  what 
will  be  the  efficiency  of  the  compensator  with  a  load  of  60  lamps, 
evenly  distributed  on  the  4  branches  ? 

24.  What  will  be  the  efficiency  of  the  compensator  with  the 
load  of  Problem  22  ? 

25.  What  is  the  efficiency  of  the  compensator  with  60  lamps 
evenly  divided  between  2  branches  ? 

26.  The  transformer  described  in  Problem  16  is  used  as  a 
booster  on  an  1100  volt  lighting  circuit.     A  lineman  thought- 
lessly removes  the  primary  fuse  at  a  time  when  there  is  a  load 
of  40   kilowatts   upon   the   circuit.     Describe   the   effects  and 
roughly   estimate   the   value   of   the    resulting   voltage    in   the 
primary  of  the  transformer. 


CHAPTER   XXIII 
THE   ROTARY    CONVERTER 

The  rotary  converter  consists  of  a  direct  current  dynamo 
with  two  or  more  collecting  rings  connected  to  symmetrical 
points  in  the  armature  winding.  Ordinarily  such  machines  are 
driven  from  the  alternating  current  side,  and  generate  direct 
current.  When  driven  by  the  direct  current  they  are  termed 
inverted  rotaries. 

In  a  single  phase  rotary  it  is  evident  that  the  constant  value 
of  the  continuous  electro-motive  force  will  equal  the  maximum 
value  of  the  alternating  electro-motive  force.  Clearly,  also,  the 
average  power  in  the  primary  will  equal  the  average  power  in 
the  secondary,  plus  the  loss  in  the  rotary.  Consider  the  case  of 
an  inverted  rotary  supplying  single  phase  current  to  a  circuit 
of  unit  power  factor;  assuming  the  secondary  electro-motive 
force  and  current  to  vary  harmonically  and  neglecting  losses, 

we  have  virtual  value  of  Ec>=— i,  since  the  maximum  value  of 

V2 

E<i  is  equal  to  E^  Hence  virtual  value  of  J2  =  V2  Iv  and  maxi- 
mum value  of  /2  =  2  1^.  Therefore  the  value  of  the  secondary 
power  must  fluctuate  between  zero  and  twice  the  primary 
power,  with  double  the  frequency  of  the  alternating  current. 

Evidently,  then,  the  angular  velocity  of  the  armature  cannot 
be  constant.  When  the  alternating  current  is  zero,  all  the 
primary  power  is  spent  in  accelerating  the  armature ;  when  the 
alternating  current  has  its  maximum  value,  half  the  secondary 
power  is  derived  directly  from  the  primary,  the  other  half  being 
obtained  from  the  energy  previously  stored  up  in  the  rotating 
armature. 

124 


THE   ROTARY  CONVERTER  125 

The  field  and  hysteresis  losses  in  a  converter  will  be  those  of 
an  alternating  dynamo,  but  the  armature  I^R  loss  will  be  greater 
in  a  single  phase  and  less  in  a  polyphase  rotary  than  those  of 
the  ordinary  dynamo. 

PROBLEMS 

1.  Neglecting   all  losses,   what  is  the  virtual  value  of  the 
armature  current  in  an  inverted  single  phase  rotary  converter, 
with  a  direct  current  input  of  50  amperes  ? 

2.  What  is  the  voltage  in  the  primary  of  a  3  phase  rotary 
supplying  direct  current  to  a  street  railway  feeder  at  550  volts  ? 

3.  If  the  direct  current  output  in  the  machine  of  Problem  2 
is  100  amperes,  what  will  be  the  primary  line  current  in  phase 
with  the  electro-motive  forces  ? 

4.  A  single  phase  2  J  kilowatt  inverted  rotary  is  supplied  with 
current  from  110  volt  lighting  mains.     With  1  ampere  in  the 
field,  the  secondary  power  factor  is  unity,  and  2  amperes  are 
required  to  drive  the  armature  at  full  speed  with  open  secondary. 
The  armature  resistance  between  primary  brushes  is  .15  ohm, 
and  the  I^R  loss  as  a  converter  is  14J&  greater  then  as  a  direct 
current   generator   giving  the   same  output  at  the  impressed 
voltage.     What  is  the  efficiency  at  full  load  ? 

5.  If  the  14^)  increase  of  I2R  loss  in  the  above  armature  at 
unit  power  factor  becomes  37^)  with  a  power  factor  of  .8,  what 
will  then  be  the  efficiency  at  full  load  ? 

6.  What  will  be  the  secondary  current  in  the  above  case  ? 

7.  Neglecting  losses,  what  is  the  virtual  armature  current  in 
a  3  phase  10  kilowatt  converter  delivering  direct  current  at 
500  volts  ? 

8.  What  is  the  speed  in  revolutions  per  minute  of  an  8  pole 
converter  running  on  a  circuit  whose  periodicity  is  60  ? 

9.  A   16  pole  multiple    drum  converter   is  driven    from  a 
3  phase,  25  cycle,  320  volt  line.     Each  pole  has  an  area  of  752 
square  centimeters,  and  the  flux  density  in  the  gap  is  8.27  kilo- 
gausses.     Determine  the  number  of  armature  conductors. 


126  ELECTRICAL  PROBLEMS 

10.  The  current  density  in  the  armature  conductors  of  the 
above  machine,  if  run  as  a  500  kilowatt  direct  current  generator, 
would  be  225  amperes  per  square  centimeter,  and  the  armature 
I^R  loss  would  be  38^>  greater  than  when  equally  loaded  as 
a  converter.     The  total   length  of  an   armature   conductor  is 
76  centimeters.     What  is  the  I2fi  loss  in  the  armature  with  a 
load  of  600  kilowatts  if  the  machine  runs  at  45°  C.  ? 

11.  Determine  the  efficiency  of  the  above  converter  at  loads 
of  200,  400,  and  600  kilowatts,  assuming  a  core  loss  of  7000 
watts,  8  horse  power  lost  in  mechanical  friction,  and  a  loss  of 
5000  watts  in  the  field. 


CHAPTER    XXIV 
THE  INDUCTION  MOTOR 

The  fundamental  equation  of  that  form  of  the  transformer 
known  as  the  induction  motor  is  E  =  V2  7rnqN<&x~LQ~s,  where 
E  is  the  virtual  value  of  the  electro-motive  force  induced  in  one 
phase  of  the  stator  or  rotor  winding,  of  N  convolutions  in  series, 
by  the  field  of  <f>  maxwells  per  pole,  rotating  at  such  velocity 
as  to  produce  the  periodicity  n ;  q  is  the  breadth  coefficient  of 
the  coils. 

Fig.  33  shows  the  flux  <I>  produced  by  the  exciting  current 
On,  the  resultant  of  the  magnetizing  component  Om  and  the 
hysteretic  component  mn.  This  flux  sweeping  round  with  the 


FIG.  33 

angular  velocity  2  TTU  induces  the  electro-motive  force  e\  in  one 
phase  of  the  stator  conductors,  and  the  electro-motive  force  e2 
in  one  phase  of  the  rotor  conductors. 

OR^   volts   of  the   electro-motive  force  e^  are  consumed  in 
setting  up  the  current  J2  through  the  rotor  resistance  R^  and 

127 


128  ELECTRICAL  PROBLEMS 

e^Rz  volts  in  overcoming  the  rotor  reactance  due  to  rotor  slip 
and  leakage. 

The  current  /2  in  the  secondary  demands  the  component  n/j 
in  the  primary,  due  allowance  being  made  for  the  ratio  of 
transformation,  and  this  compounded  with  the  exciting  current 
On  gives  the  primary  current  OIr  To  establish  this  current 
demands  OJK1  volts  to  overcome  stator  resistance,  and  R±L  volts 
to  overcome  the  reactance  due  to  stator  leakage.  The  electro- 
motive force  OL  required  to  overcome  impedance  compounded 
with  Oev  that  required  to  overcome  the  counter  electro-motive 
force  Oe'v  gives  the  impressed  electro-motive  force  OEr  The 
reactance  due  to  magnetic  leakage  in  stator  and  rotor  may  be 
experimentally  determined,  or  it  may  be  approximately  calcu- 
lated by  estimating  the  reluctance  of  the  leakage  path  between 
adjacent  teeth,  and  hence  the  maximum  value  of  the  leakage 
flux  encircling  a  slot  when  a  current  whose  virtual  value  is 
unity  flows  through  N'  conductors  in  that  slot.  This  should 
be  increased  by  20^  more  or  less  on  account  of  the  flux 
encircling  the  end  connections.  The  total  leakage  flux  divided 
by  108  is  the  inductance  in  henrys  per  conductor;  this  multi- 
plied by  the  number  of  conductors  in  one  phase  gives  the 
inductance  per  phase. 

From  a  knowledge  of  the  total  power  imparted  to  the  rotor, 
the  rotor  current  may  be  calculated  from  the  equation  watts 
=  n!NzI2<&  X  13.28  X  10~8,  where  JV2  is  the  number  of  convolu- 
tions per  phase  of  the  rotor  winding  and  n'  =  (1— s)w,  s  being 
the  per  cent  of  slip.  The  above  equation  is  derived  from  fun- 
damental principles  on  the  assumption  of  the  sine  distribution 
of  currents  and  flux  about  the  perimeter  of  the  rotor. 

In  calculating  the  magnetizing  component  of  the  exciting 
current,  it  should  be  borne  in  mind  that  on  the  assumption  of 
sine  distribution  of  /  and  <E>  the  ratio  of  the  resultant  field  to 
the  field  produced  by  one  phase  is  half  the  number  of  phases. 

In  the  following  problems  the  construction  of  the  clock 
diagram  should  proceed  along  with  the  numerical  calculations. 


THE  INDUCTION  MOTOK  129 

PROBLEMS 

1.  A  6  pole  5  horse  power  induction  motor  is  supplied  with 
3  phase  current  at  110  volts  and  a  periodicity  of  60.    The  stator 
phases  are  connected  in  mesh,  each  phase  of  the  winding  being 
equivalent  to  18  coils  of  10  convolutions  of  No.  9  wire,  B.S.G., 
each.     The  stator  has  54  slots  .95  centimeter  wide  and  3.8  cen- 
timeters deep;  the  rotor  has  72  slots  .8  centimeter  wide  and 
3.2   centimeters  deep.     The  rotor  phases  are  star  connected  and 
joined  to  3  collecting  rings,  each  phase  consisting  of  48  bars  of 
copper  1.27  centimeters  deep  and  .475  centimeter  wide.     The 
length  of  laminations  parallel  to  the  shaft,  excluding  ventilat- 
ing ducts,  is  11.5  centimeters ;  the  depth  of  iron  under  slots 
is   2.3    centimeters    in   both    stator  and  rotor.      The   external 
diameter  of  the  rotor  is  35.5  centimeters,  and  the  air  gap  is 
.125  centimeter.     What  will  be  the  synchronous  speed  of  the 
motor  ? 

2.  Assuming  10%  drop  in  the  stator  and  a  breadth  coefficient 
of  .95,  what  is  the  value  of  the  flux  per  pole  ? 

3.  Assuming  a  sine  distribution  of  the  flux  over  the  polar 
area  and,  owing  to  the  spread  of  the  lines  from  the  sides  of  the 
teeth,  assuming  the  effective  polar  area  to  be  .6  of  the  polar 
pitch  multiplied  by  the  length  of  laminations,   calculate  the 
magnetizing  component  of  the  exciting  current  in  each  phase 
of  the  stator  winding. 

4.  Calculate  the  hysteretic  component  of  the  exciting  current, 
assuming  a  hysteretic  coefficient  of  .003. 

5.  Calculate  the  resistance  per  phase  of  the  rotor  at  40°  C., 
the  virtual  length  of  each  bar  being  40.6  centimeters. 

6.  Assuming  the  loss  by  mechanical  friction  to  be  .25  horse 
power,  assuming  5%  slip,  and  neglecting  the  I\R^  loss,  calculate 
the  rotor  current  at  full  load. 

7.  Recalculate  the  rotor  current,  adding  the  probable  /22AJ2 
loss,  as  deduced  from  the  approximate  value   of  the   current 
above  calculated,  to  the  rotor  loss. 


130  ELECTEICAL  PROBLEMS 

8.  At  what  per  cent  slip  will  the  above  rotor  current  be 
generated  ? 

9.  Calculate  the  self-inductance  of  a  phase  of  the  rotor  due 
to  magnetic  leakage,  assuming  the  leakage  in  the  slot  to  extend 
down  1.2  centimeters  from  the  surface  of  the  rotor. 

10.  Calculate  the  rotor  reactance. 

11.  Calculate   the   lag  angle   of   the  rotor    current   at    full 
load. 

12.  Determine   the  ratio  of   transformation   and  thence  the 
rotor  component  of  the  stator  current. 

13.  Determine  the  total  full  load  stator  current. 

14.  Assuming  the  mean  length  of  a  convolution  of  the  stator 
winding  to  be  82  centimeters,  calculate  the  stator  resistance 
at  50°  C. 

15.  Determine  the  /21^1  loss  in  the  stator. 

16.  Calculate  the  reactance  of  the  stator  winding,  assuming 
the  leakage  in  the  slot  to  extend  inward  1.6  centimeters  from 
the  end  of  the  tooth. 

17.  Complete  the  clock  diagram,  determining  the  exact  value 
of  the  impressed  electro-motive  force. 

18.  Determine  the  full  load  power  factor  from  the  diagram, 
and  also  from  the  ratio  of  the  real  and  apparent  powers. 

19.  Determine  the  full  load  efficiency. 

20.  Calculate  the  no  load  stator  current  and  power  factor, 
first  roughly  estimating  the  primary  drop  and  slip. 

21.  Draw  the  complete  clock  diagram  for  the  above  motor  at 
\  full  load  and  determine  the  power  factor. 

22.  Draw  the  complete  clock  diagram  for  the  above  motor  at 
3-  full  load  and  determine  the  power  factor. 

23.  Draw  the  complete  clock  diagram  for  the  above  motor  at 
f  full  load  and  determine  the  power  factor. 

24.  What  are  the  line  currents  at  each  of  the  above  loads  ? 

25.  What  resistance  must  be  inserted  in  each  phase  of  the 
rotor  circuit  in  order  that  it  shall  exert  full  load  torque  when 
starting  with  full  load  current  ? 


THE   INDUCTION  MOTOR  131 

26.  If  the  above  stator  had  been  made  with  48  slots,  and 
the  supply  had  been  2  phase  at  156  volts,  the  breadth  coefficient 
being  .91,  and  all  conditions  not  necessarily  altered  by  these 
changes  remaining  the  same,  what  would  have  been  the  stator 
current  at  each  quarter  of  full  load  ? 

27.  In  the  case  of  the  3  phase  motor  treated,  draw  curves 
with  output  in  horse  power  as  abscissas  and  with  stator  currents, 
slips,   efficiencies,  torques,  and  power  factors,  respectively,  as 
ordinates. 

28.  Draw  curves  similar  to  those  of  Problem  27  for  the  2  phase 
motor  treated. 


CHAPTER   XXV 
TESTING   OF   DYNAMOS 

One  of  the  most  satisfactory  methods  of  determining  the 
efficiency  of  dynamos  consists  in  the  observation  of  the  no  load 
armature  losses  due  to  brush  and  bearing  friction,  hysteresis, 
eddy  currents,  and  air  friction.  Of  these  the  first  two  vary 
directly  as  the  speed,  the  tliird  as  the  square  of  the  speed,  and 
the  fourth  as  the  cube  of  the  speed.  With  a  given  strength  of 
field,  we  may  accordingly  write 


armature  losses  =  I^R  +  (A  +  a)  n  +  En*  +  Oi3, 

where  n  is  the  speed,  A,  a,  B,  and  C  being  constants  for  the 
particular  machine  and  strength  of  field.  Moreover  A  and  C 
are  evidently  independent  of  the  field,  while  a  varies  as  the 
1.6  power  and  B  as  the  square  of  the  field  strength. 

The  relative  value  of  the  flux  with  different  field  currents 
will  evidently  vary  inversely  with  the  speeds  running  as  a  motor 
without  load  at  a  constant  armature  electro-motive  force,  allow- 
ance being  made,  if  necessary,  for  the  slight  IE  drop  in  the 
armature  conductors  themselves. 

To  determine  the  constants  J,  a,  B,  and  C,  run  the  machine 
as  a  motor  without  load  with  a  constant  current  in  the  field, 
varying  the  speed  by  altering  the  impressed  voltage.  From  the 
data  thus  obtained  construct  a  curve  with  armature  watts,  minus 
J2j£,  as  ordinates  and  speed  in  revolutions  per  minute  as  abscissas. 
Draw  a  tangent  to  this  curve  through  the  origin,  and  the  ordi- 
nate  of  this  tangent  for  any  particular  speed  will  give  the  loss 
due  to  brush  and  bearing  friction  and  hysteresis.  In  like  man- 
ner construct  one  or  two  other  curves  and  tangents,  each  Avith 

132 


TESTING  OF  DYNAMOS  133 

a  constant  but  widely  different  field  current.  The  values  of  A 
and  a  can  now  be  found,  remembering  that  A  is  the  same  for 
all  curves  while  a  varies  as  the  1.6  power  of  the  field  strength. 
B  and  C  can  be  easily  determined  analytically.  The  values  of 
the  constants  thus  obtained  should  be  verified  by  seeing  that 
they  give  points  on  the  curves  within  the  limits  of  the  allowable 
errors  of  observation. 

Finally  the  constants  a  and  B  may  be  changed  to  h<&l-Q  and 
#<l>2,  where  h  and  e  are  respectively  the  hysteresis  and  eddy 
current  constants  and  <I>  is  the  flux  in  megamaxwells. 

Knowing  the  armature  resistance,  the  total  armature  losses  for 
any  load  may  now  be  calculated  with  considerable  precision. 

Railway  motors  are  generally  tested  by  mounting  the  axle 
gears  of  two  similar  machines  upon  a  common  shaft,  the  second 
machine  being  driven  as  a  generator  by  the  first.  The  efficiency 
of  the  combination  is  of  course  the  ratio  of  the  generator  output 
to  the  motor  input.  When  the  resistance  drop  in  the  two  machines 
is  not  excessive,  it  is  customary  to  consider  the  efficiency  of  the 
motor  as  the  square  root  of  the  efficiency  of  the  combination. 

The  student  can  easily  acquaint  himself  with  the  other 
methods  of  testing  referred  to  in  the  following  problems. 

PROBLEMS 

1.  An  arc  light  machine  was  mounted  on  a  Bracket  cradle 
dynamometer.    When  driven  at  800  revolutions  per  minute,  the 
output  was  observed  to  be  6.8  amperes  at  2000  volts,  and  equi- 
librium was  produced  by  a  weight  of  57  pounds  at  the  extremity 
of  a  3  foot  arm.     What  was  the  efficiency  of  the  machine  ? 

2.  A  toy  motor  was  tested  with  a  brake  consisting  of  a  piece 
of  twine  passing  over  a  groove  in  the  edge  of  a  5  centimeter 
brass  disk  attached  to  the  end  of  the  shaft.     To  one  end  of  the 
string  a  50  gram  weight  was  hung,  the  other  end  being  tied 
to  a  spiral  spring  carrying  an  index.     The  speed  was  determined 
by  passing  the  point  of  a  quill  attached  to  the  prong  of  a  vibrating 


134  ELECTRICAL   PROBLEMS 

tuning  fork  from  the  center  to  the  periphery  of  the  smoked 
outer  face  of  the  disk  and  counting  the  undulations  in  one  or 
more  turns  of  the  spiral  thus  produced. 

The  following  observations  were  made:  rate  of  fork  64 
periods  per  second  with  16  complete  periods  in  5  revolutions ; 
extension  of  spiral  spring  =  force  of  78  grams ;  input  of  motor 
5  amperes  at  4  volts.  What  was  the  efficiency  of  the  motor? 

3.  A  small  motor  was  tested  with  a  brake  consisting  of  a 
cord  and  a  6  inch  grooved  pulley.      One  end  of  the  cord  was 
attached  to  a  5  pound  weight  resting  in  the  pan  of  a  grocer's 
balance ;  to  the  other  end  was  fastened  a  1  pound  weight  hang- 
ing free.     When  running  at  1800  revolutions  per  minute,  the 
apparent  weight  in  the  pan  was  observed  to  be  2  pounds,  the 
input  of  the  motor  being  4  amperes  at  100  volts.     What,  was 
the  efficiency  of  the  motor? 

4.  A  shunt  wound  2  pole  dynamo  has  an  armature  of  .06  ohm 
resistance.     The  machine  was  run  as  a  motor  without  load,  and 
from  the  following  observations  were  plotted   3   curves,  each 
with  a  different  strength  of  field,  with  revolutions  per  minute 
as  abscissas  and  watts  delivered   to  armature,  minus  I^R  in 
armature,  as  ordinates. 

WATTS  IN  ARMATURE  MINUS   F2R  IN  ARMATURE 

Speed,  R.P.M.  Field  Current  Field  Current  Field  Current 

('}.'•>  Amperes)  (5.0  Amperes)  (7.5  Amperes) 

100  22  35  46 

200  46  73  96 

300  71  115  152 

400  98  160  213 

500  127  210  281 

600  15!)  264  355 

700  194  324  436 

800  232  388  523 

900  274  459  618 

1000  320  534  722 

1100  370  616  833 

1200  425  705  953 


TESTING  OF  DYNAMOS  135 

The  relative  values  of  the  total  flux  with  different  currents 
in  the  field  were  obtained  from  the  following  observations : 

Volts  at  Brushes  Amperes  in  Field  Revolutions  per  Minute 
60                                          2.5  1220 

(50  5.0  660 

60  7.5  510 

Determine  the  constants  A,  a,  7>,  (7  for  each  of  the  above  field 
currents. 

5.  Calculate  the  armature  losses  of  the  above  machine  running 
as  a  generator  at  1150  revolutions  per  minute  with  7.5  amperes 
in  the  field  and  an  external  load  of  50  amperes,  there  being  no 
lead  to  the  brushes. 

6.  Calculate  the  efficiency  of  the  above  machine  when  running 
at  1025  revolutions  with  5  amperes  in  the  field,  there  being 
80  volts  at  the  brushes  and  70  amperes  in  the  external  circuit. 

7.  A  small  3  phase  alternator  was  motor  driven  by  the  dynamo 
described  in  Problem  4  at  1200  revolutions  per  minute,  when 
the  input  of  the  motor  armature  was  observed  to  be  50  amperes 
at  150  volts  and  the  field  current  7.5  amperes.     A  wattmeter  in 
one  branch  of  the  3  phase  circuit  showed  1800  watts,  and  the 
alternator  fields  required  4  amperes  at  100  volts.      What  was 
the  efficiency  of  the  alternator? 

8.  A  2  pole  dynamo  of  the  Sprague  motor  type  has  each  of 
its  fields  wound  with  1750  convolutions  of  wire,  the  2  coils  being 
connected  in  series.     There  are  44  coils,  each  of  2  convolutions, 
upon  the  armature,  and  with  97  volts  at  the  brushes  it  was  run  as 
a  motor  without  load  at  880  revolutions  per  minute,  the  field 
current  being  7.2  amperes.  With  the  same  voltage  and  resistance 
in  the  field  circuit,  the  following  observations  were  made : 

Time  in  Seconds  from 

closing  Field  Circuit  Current  Amperes 

1.0  4.0 

2.5  6.2 

4.5  7.0 

5.5  7.1 

8      and  over  7.2 


136  ELECTRICAL  PROBLEMS 

Determine  the  time  constant  of  the  shunt  field  circuit  and 
hence  the  total  flux  through  each  leg  of  the  field  magnet.  From 
observed  voltage  and  speed  and  the  given  armature  winding, 
determine  the  flux  through  the  armature  and  hence  the  leakage 
coefficient. 

9.  The  following  observations  were  taken  upon  a  pair  of 
street  railway  motors  whose  axle  gears  were  mounted  upon  a 
common  shaft,  the  ratio  of  reduction  being  4.78.  Motor  No.  1 
was  run  by  current  from  the  street  railway  mains ;  motor  No.  2 
was  caused  to  generate,  its  output  being  absorbed  by  a  large 
rheostat.  Each  motor  had  a  resistance  of  1.1  ohms. 

MOTOR 

Volts  Amperes 
521  20.1 

525  21.0 

518  25.4 

517  27.3 

514  32.0 

518  40.2 

515  50.0 

516  60.0 
516  72.8 
513  75.8 

Reduce  the  observations  to  a  basis  of  a  constant  voltage  of 
500,  the  speed  being  proportional  to  the  voltage,  and  from  the 
observations  construct  the  following  curves : 

1.  A  speed  curve  having  current  in  amperes  for  ordinates 
and  a  double  set  of  abscissas,  one  revolutions  per  minute  of  the 
armature  and  the  other  speed  in  miles  per  hour  of  a  car  with 
30  inch  wheels. 

2.  A    torque    curve    having    current    for    ordinates    and    a 
double    set   of    abscissas,   one   car  axle    torque   and   the   other 
tractive  effort. 

3.  A    curve   having  current   as   ordinates   and   efficiency  as 
abscissas. 


GENERATOR 

Armature  Speed, 

Volts 

Amperes 

R.P.M. 

393 

15.2 

789 

373 

14.8 

768 

389 

19.7 

690 

392 

21.6 

663 

401 

25.8 

612 

382 

34.8 

547 

360 

44.0 

488 

341 

52.1 

445 

308 

65.2 

404 

302 

67.8 

393 

TESTING  OF  DYNAMOS  137 

10.  A  pair  of  500  volt  railway  motors  of  .75  ohrn  resistance 
each  were  tested  in  the  usual  way,  the  following  readings  being 
taken : 

MOTOR 

Volts  Amperes 

515  9.8 

513  20.5 

520  26.3 

512  31.2 

509  36.1 

509  46.8 

506  56.5 

505  68.3 

490  100.5 

Draw  the  500  volt  speed  curve,  the  reduction  ratio  being  4.78, 
and  the  curve  of  tractive  effort  with  30  inch  wheels.  From 
these  construct  curves  of  efficiency  as  depending  on  load  and 
on  speed. 


GENERATOR 

Volts             Amperes 

Armature  Speed, 
R.P.M. 

50 

0.0 

887 

363 

12.8 

525 

386 

17.6 

497 

388 

20.3 

473 

398 

25.1 

454 

388 

30.7 

435 

377 

37.9 

412 

363 

45.1 

398 

317 

58.4 

383 

CHAPTER    XXVI 
THE   TRANSMISSION   OF  POWER 

Power  is  now  transmitted  electrically  at  pressures  up  to  about 
60,000  volts,  and  to  distances  considerably  above  100  miles. 
Reactances  interfere  with  regulation,  and  by  lowering  the  power 
factor  diminish  the  efficiency  of  the  transmission,  but  both  theory 
and  experience  show  that  only  the  I2R  losses  are  of  material 
importance. 

The  student  should  carefully  examine  the  solution  of  each 
of  the  following  problems,  and  enumerate  such  general  prin- 
ciples as  he  may  be  able  to  deduce  therefrom.  Unless  otherwise 
specified,  unit  power  factor,  a  line  temperature  of  15°  C.,  and  in 
the  case  of  copper  a  conductivity  97^  that  of  the  pure  metal, 
will  be  assumed. 

In  the  case  of  polyphase  transmission  it  is  well  to  carry  on 
the  computations  with  reference  to  a  single  wire  of  the  line. 
Thus  in  three  phase  lines  the  voltage  per  line  is  E -%-  V3;  the 
resistance,  reactance,  and  impedance  per  unit  length  will  each  be 
half  that  of  the  two  wires  in  a  single  phase  line. 

PROBLEMS 

1.  Determine  the  weight  of  copper  per  kilowatt  delivered  by 
a  10,000  volt  generator  over  a  transmission  line  10  miles  long, 
at  efficiencies  varying  from  10%  to  95%.     Construct  a  curve 
with  efficiencies  for  abscissas  and  weights  of  copper  per  kilo- 
watt delivered  as  ordinates. 

2.  Construct    a    curve    showing  the  weight  of   copper  per 
kilowatt  delivered  at  an  efficiency  of  90%  by   a  10,000  volt 

138 


-  THE  TRANSMISSION  OF  POWER  139 

generator    over    a   transmission   line   of   length  varying  from 
1  mile  to  100  miles. 

3.  Construct  a  curve  showing  the  weight  of  copper  per  kilowatt 
delivered  at  an  efficiency  of  90%  over  a  transmission  line  10  miles 
long,  with  generated  voltages  varying  from  1000  to  10,000. 

4.  If  power  at  the  generator   costs   $20   per  kilowatt  per 
annum  and  copper  costs  18  cents  per  pound,  how  far  will  it  pay 
to  transmit  power  from  a  10,000  volt  generator  over  a  line  of 
80%  efficiency  to  a  point  where  the  selling  price  of  power  is 
$50  per  kilowatt  per  annum?     Assume  the  total  cost  of  the 
line  to  be  twice  that  of  the  copper  in  it  and  allow  10%  of  the 
cost  of  the  line  annually  for  interest  and  depreciation. 

5.  What  will  be  the  value  of  power  delivered  at  30,000  volts 
over  a  75  mile  line  at  85%  efficiency,  with  power  at  the  gen- 
erator costing  $15  per  kilowatt  per  year  and  with  the  same 
assumptions  as  to  cost  of  copper,  total  cost  of  line,  and  interest 
and  depreciation  as  in  Problem  4  ? 

6.  What  is  the  voltage  of  the  generator  when  power  is  deliv- 
ered at  10,000  volts  and  a  periodicity  of  30  to  a  non-reactive 
load  of  200  ohms  effective  resistance,  over  a  line  10  miles  long, 
of  No.  0  B.S.G.  aluminum  wires  hung  30  inches  apart?    Assume 
the  distributed  capacity  of  the  line  to  be  equivalent  to  that  of 
an  equal  capacity  shunted  across  the  middle  of  the  line. 

7.  Assume  the  length  of  the  line  in  Problem  6  to  be  increased 
to  25  miles,  the  terminal  voltage  to  20,000,  and  the  effective 
resistance  of  the  non-reactive  load  to  400  ohms.    What  will  then 
be  the  generator  voltage  ? 

8.  Assume  the  length  of  line  in  Problem  6  to  be  increased  to 
100  miles,  the  terminal  voltage  to  40,000,  and  the  resistance 
of  the  non-reactive  load  to  800  ohms.     What  will  then  be  the 
generator  voltage? 

9.  Assume  the  length  of  line  in  Problem  6  to  be  increased  to 
150  miles,  the  terminal  voltage  to  60,000,  and  the  resistance  of 
the  non-reactive  load  to  1200  ohms.     What  will  then  be  the 
generator  voltage? 


140  ELECTRICAL  PROBLEMS 

10.  What  would  have  been  the  generator  voltage  in  Problem  9 
if  we  had  assumed  the  distributed  capacity  of  the  line  to  have 
been  replaced  by  three  condensers,  viz.,  one  at  each  end  of  the 
line  of  £  the  line  capacity  and  one  across  the  middle  of  the  line 
of  |  the  line  capacity? 

11.  What  is  the   maximum  non-reactive  load  that  can  be 
delivered  over  the  line  of  Problem  8  ? 

12.  If  500  kilowatts  are  delivered  at  10,000  volts  over  the 
line  described  in  Problem  6,  neglecting  the  line  capacity,  to  an 
overexcited  synchronous  motor  with  a  power  factor  of  .8,  from 
generating  machinery  of  14.22  ohms  resistance  and  .474  henry 
inductance,  what  is  the  generated  electro-motive  force  ? 

13.  With  the  system  of  Problem  12,  draw  a  curve  showing 
the  electro-motive  force  of  the  generator  at  each  quarter  of  full 
load  from  no  load  to  50%  overload,  the  voltage  delivered  being 
10,000  in  all  cases. 

14.  If  in  Problem  13  we  neglect  the  capacity  of  the  line 
and  the  resistance  and  inductance  of  the  generating  machinery, 
what  is  the  highest  power  factor  at  which  500  kilowatts  can  be 
delivered  without  drop  of  voltage  on  the  line  ? 

15.  With  a  given  maximum  voltage  between  lines,  what  is 
the   relative   copper    economy  of   single   phase,   2   phase  with 
common  return,  and  3  phase  transmission  lines? 

16.  What  will  be  the  diameter  of  copper  wires  needed  to 
deliver  5000  horse  power  at  a  distance  of  20  miles  with  a  work- 
ing pressure  of  20,000  volts  and  a  power  factor  of  .9,  over  a 
3  phase  transmission  line  at  an  efficiency  of  85  %  ? 

17.  A  3  phase  line  250  miles  long  consists  of  3  No.  000  hard 
drawn  copper  wires  arranged  in  the  form  of  a  triangle,  15  feet 
on  a  side.    At  the  end  of  this  line  10,000  horse  power  are  deliv- 
ered at  a  pressure  of  70,000  volts  and  a  periodicity  of  20,  to  an 
inductive  load  with  a  power  factor  of  .9.    Treating  the  capacity 
of  the  line  as  in  Problem  10  and  assuming  leakage  to  be  absent, 
what  is  the  efficiency  of  the  transmission  and  what  the  power 
factor  at  the  generator? 


APPENDIX 


WIRE   TABLE,  ANSWERS,  AND   GRAPHICAL 
SOLUTIONS 


142 


ELECTEICAL  PROBLEMS 
WIRE   TABLE 


No. 

DIAMETER 

AREA 

WEIGHT 

B.S.G. 

mils 

mm. 

circ.  mils 

sq.  mm. 

Ibs.  per 
1000ft. 

kgs.  per 
km. 

0000 

460 

11.684 

211,600 

107.2 

640.5 

953.1 

000 

410 

10.414 

168,100 

85.18 

508.9 

757.2 

00 

365 

9.271 

133,225 

67.52 

403.3 

600.1 

0 

325 

8.255 

105,625 

53.18 

319.7 

475.8 

1 

289 

7.341 

83,521 

42.33 

252.8 

376.2 

2 

258 

6.553 

66,564 

33.73 

201.5 

299.8 

3 

229 

5.827 

52,441 

26.73 

158.7 

236.2 

4 

204 

5.182 

41,616 

21.10 

126.0 

187.5 

5 

182 

4.623 

33,124 

16.79 

100.3 

149.2 

6 

162 

4.115 

26,244 

13.30 

79.44 

118.2 

7 

144 

3.658 

20,736 

10.51 

62.77 

93.40 

8 

128 

3.251 

16,384 

8.303 

49.48 

73.80 

9 

114 

2.896 

12,996 

6.588 

39.34 

58.54 

10 

102 

2.591 

10,404 

5.274 

31.49 

46.86 

11 

91 

2.311 

8,281 

4.195 

25.10 

37.35 

12 

81 

2.057 

6,561 

3.324 

19.86 

29.55 

13 

72 

1.829 

5,184 

2.628 

15.69 

23.35 

14 

64 

1.626 

4,096 

2.077 

12.40 

18.45 

15 

57 

1.448 

3,249 

1.647 

9.972 

14.84 

16 

51 

1.295 

2,601 

1.317 

7.873 

11.72 

17 

45 

1.143 

2,025 

1.026 

6.130 

9.121 

18 

40 

1.016 

1,600 

.8109 

4.843 

7.207 

19 

36 

.9144 

1,296 

.6568 

3.923 

5.838 

20 

32 

.8130 

1,024 

.5193 

3.100 

4.613 

21 

28.5 

.7239 

812.3 

.4117 

2.459 

3.659 

22 

25.3 

.6426 

640.1 

.3244 

1.938 

2.883 

23 

22.6 

.5740 

510.8 

.2588 

1.546 

2.301 

24 

20.1 

.5105 

404.0 

.2047 

1.223 

1.820 

25 

17.9 

.4547 

320.4 

.1624 

.9699 

1.443 

26 

15.9 

.4039 

252.8 

.1282 

.7652 

1.139 

27 

14.2 

.3607 

201.6 

.1022 

.6103 

.9081 

28 

12.6 

.3200 

158.8 

.0804 

.4807 

.7153 

29 

11.3 

.2870 

127.7 

.0647 

.3866 

.5752 

30 

10 

.2540 

100.0 

.0507 

.3027 

.4504 

31 

8.9 

.2261 

79.21 

.0402 

.2398 

.3568 

32 

8 

.2032 

64.00 

.0324 

.1938 

.2883 

33 

7.1 

.1803 

50.41 

.0255 

.1526 

.2271 

34 

6.3 

.1600 

39.69 

.0201 

.1201 

.1788 

35 

5.6 

.1422 

31.36 

.0159 

.0949 

.1413 

36 

5 

.1270 

25.00 

.0127 

.0757 

.1124 

WIRE   TABLE 


143 


WIRE   TABLE 


RESISTANCE 


o°c. 

20°  C. 

40°  C. 

60°  C. 

1000ft. 

km. 

1000  ft. 

km. 

1000  ft. 

km. 

1000ft. 

km. 

.0452 

.1482 

.0488 

.1601 

.0524 

.1719 

.0560 

.1838 

.0569 

.1866 

.0614 

.2015 

.0660 

.2164 

.0705 

.2314 

.0718 

.2354 

.0775 

.2543 

.0832 

.2731 

.0890 

.2919 

.0905 

.2969 

.0978 

.3207 

.1050 

.3445 

.1122  ' 

.3682 

.1145 

.3755 

.1236 

.4056 

.1328 

.4356 

.1419 

.4657 

.1436 

.4712 

.1551 

.5089 

.1666 

.5466 

.1781 

.5843 

.1823 

.5981 

.1969 

.6459 

.2115 

.6938 

.2261 

.7416 

.2297 

.7537 

.2481 

.8140 

.2665 

.8743 

.2849 

.9346 

.2886 

.9469 

.3117 

1.023 

.3348 

1.098 

.3579 

1.174 

.3643 

1.195 

.3934 

1.291 

.4226 

1.386 

.4517 

1.482 

.4610 

1.513 

.4979 

1.634 

.5348 

1.755 

.5717 

1.876 

.5835 

1.915 

.6302 

2.068 

.6769 

2.221 

.7235 

2.374 

.7356 

2.413 

.7945 

2.606 

.8533 

2.800 

.9122 

2.993 

.9189 

3.015 

.9924 

3.256 

1.066 

3.497 

1.139 

3.738 

1.154 

3.788 

1.247 

4.090 

1.339 

4.394 

1.432 

4.697 

1.457 

4.781 

1.574 

5.163 

1.690 

5.545 

1.807 

5.928 

1.844 

6.050 

1.992 

6.534 

2.139 

7.018 

2.287 

7.502 

2.334 

7.657 

2.520 

8.270 

2.707 

8.883 

2.894 

9.495 

2.942 

9.654 

3.178 

10.43 

3.413 

11.20 

3.649 

11.97 

3.676 

12.06 

3.970 

13.02 

4.264 

13.99 

4.454 

14.95 

4.721 

15.49 

5.099 

16.73 

5.476 

17.97 

.  5.854 

19.21 

5.975 

19.60 

6.453 

21.17 

6.931 

22.74 

7.409 

24.31 

7.377 

24.20 

7.967 

26.14 

8.557 

28.07 

9.147 

30.01 

9.336 

30.63 

10.08 

33.08 

10.86 

35.53 

11.58 

37.98 

11.77 

38.61 

12.71 

41.70 

13.65 

44.79 

14.59 

47.88 

14.94 

49.00 

16.13 

52.92 

17.32 

56.89 

18.52 

60.76 

18.72 

61.40 

20.21 

66.32 

21.71 

71.23 

23.20 

76.14 

23.66 

77.64 

25.56 

83.85 

27.45 

90.06 

29.34 

96.27 

29.84 

97.89 

32.22 

105.7 

34.61 

113.6 

37.00 

121.4 

37.82 

124.1 

40.84 

134.0 

43.87 

143.9 

46.89 

153.8 

47.42 

155.6 

51.21 

168.0 

55.01 

180.5 

58.80 

192.9 

60.20 

197.5 

65.02 

213.3 

69.83 

229.1 

74.65 

244.9 

74.86 

245.6 

80.85 

265.3 

86.84 

284.9 

92.83 

304.6 

96.00 

313.7 

103.2 

338.7 

110.9 

363.8 

118.5 

388.9 

120.7 

396.0 

130.3 

427.6 

140.0 

459.3 

149.6 

491.0 

149.4 

490.1 

161.3 

529.3 

173.3 

568.5 

185.2 

607.7 

189.6 

622.2 

204.8 

672.0 

220.0 

721.7 

235.2 

771.5 

240.9 

790.2 

260.1 

853.5 

279.4 

916.7 

298.7 

979.9 

304.8 

1000 

329.2 

1080 

353.6 

1160 

378.0 

1240 

382.4 

1255 

413.0 

1355 

443.6 

1455 

474.2 

1556 

ANSWERS 


1.  30.48  centimeters.  8. 

2.  1609.34  meters. 

3.  185,319  centimeters.  9. 

4.  .4115  centimeter. 

5.  81,713    square    mils,  10. 

10,40^  circular 

mils.  11. 

6.  21 1,600  circular  mils.  12. 

7.  1.016  millimeters.  13. 


CHAPTER   I 

77.244  cubic  centi- 
meters. 

506.7  square  milli- 
meters. 

8154.4  cubic  centi- 
meters. 

15.43  grains. 

437.5  grains. 

.672  pound. 


14.  1.49  kilograms. 

15.  62.43  pounds. 

16.  31.53  pounds. 

17.  6494.9  centimeters. 

18.  20,626  feet. 

19.  10.24  pounds. 

20.  262  pounds. 

21.  2.02  pounds. 


1.  2  amperes. 

2.  .25  ampere. 

3.  5  ohms. 

4.  .25  ohm. 

5.  16  volts. 

6.  30  volts. 

7.  11  ohms. 

8.  8.45  volts. 

9.  40  amperes. 


CHAPTER   II 

10.  25  amperes. 

11.  100  volts. 

12.  4  volts. 

13.  20  ohms. 

14.  57.5  ohms. 

15.  .75  ampere. 

16.  12  volts. 

17.  .19  ampere. 

18.  54.17  ohms. 


19.  100  ohms. 

20.  .000002  ampere. 

21.  52  volts. 

22.  3  amperes. 

23.  .00394  ampere. 

24.  70  amperes. 

25.  22,104  amperes. 


1.  .00000159. 

2.  9.56. 

3.  35.5  ohms. 

4.  989.34  feet. 

5.  10%. 

6.  .000037  ohm. 

7.  .0000783  ohm. 

8.  .326  ampere. 


CHAPTER   III 

9.  28.52%. 

10.  .00000965. 

11.  57.98. 

12.  .2665. 

13.  .006184  ohm. 

14.  1.19  ohms. 

15.  .1146  ohm. 

16.  1592.4  centimeters. 

145 


17.  .195  centimeter. 

18.  .1377  centimeter. 

19.  .0337  ampere. 

20.  .0000029. 

21.  17.44. 

22.  1.8  ohms. 


146  ELECTRICAL  PROBLEMS 


CHAPTER   IV 


1.  4  amperes.  8.  78  volts.  15.  .346  ampere. 

2.  33  ohms.  9.  8  amperes.  16.  .32  ampere. 

3.  1024  volts.  10.  33i%.  17.  .292  ampere. 

4.  .00005  ampere.  11.  .4  ampere.  18.  .8  ampere. 

5.  .4  ampere.  12.  .2  ampere.  19.  .5  ohm. 

6.  6.875  amperes.  13.  .133  ampere.  20.  42.5  volts. 

7.  10  amperes.  14.  .0667  ampere. 

21.  A-B,  .058  ohm;  A-C,  .111   ohm;   A-D,  .110  ohm;   B-C,  .054  ohm 

B-D,   .054    ohm;    C-D,   .111   ohm;    B  connected    to    the    commor 

"in  n  />+ irvr* 


22. 
23. 

junction. 
.892  ampere. 
.359  ampere. 

24. 
25. 

.374 
.291 

ampere, 
ampere. 

26. 
27. 

.191 

See 

ampere, 
page  157. 

CHAPTER   V 

1.  5  mhos,  ,2  ohm.  4.  .8  ampere;  12  volts. 

2.  2.05  mhos,  .488  ohm.  5.  1  ampere ;  2  amperes. 

3.  284  volts.  6.  B/A. 

7.  1.904  amperes.  10.  8  cells.  13.  .218  ampere. 

8.  33i%.  11.  .5  ampere.  14.  .177  ampere. 

9.  33^%.  12.  .38  ampere.  15.  .0909  ampere. 
16.  .0457  ampere.                                   17.  .55  ampere,  all  in  series. 

18.  1.1  amperes  ;  2  groups  in  parallel,  each  of  10  in  series. 

19.  2.2  amperes ;  4  groups  in  parallel,  each  of  5  in  series. 

20.  2.75  amperes ;  5  groups  in  parallel,  each  of  4  in  series. 

21.  5.5  amperes ;  10  groups  in  parallel,  each  of  2  in  series. 

22.  11  amperes;  all  in  parallel. 

23.  Series  45.45  amperes  ;  multiple  24.39  amperes. 

24.  1  ohm.  31.  Decreased  37 

25.  .025  ohm.  32.  Sl  =  4.004  ohms, 

26.  .00125  ampere.  sz  =  40.77  ohms, 

27.  .101  ohm.  SB  =  493.33  ohms, 

28.  Main  current  increased  98%;  gal-  r±  =  36.36  ohms, 

vanometer  current   decreased  r2  =  403.64  ohms, 

98%.  rz  =  3556  ohms. 

29.  9.9  ohms. 

30.  333.33  ohms  ;  30.303  ohms  ;  3.003 

ohms. 


ANSWERS  147 


CHAPTER   VI 

2.  20.76  volts.  4.  176  volts  at  7th  car. 

^,_2j2&-miles.  5.  474.23  volts  at  4th  car. 

6.  24  volts,  generator  and  line  ;  25  volts,  lamps. 

7.  2528  volts  generated  with  line  at  60°;  2502  volts  generated  with  line 

atO°. 

8.  9.69  volts  drop  to  10th  group.       13.  533.87  volts  at  station. 

9.  11.98  volts  drop  to  10th  group.     14.  468.04  volts  at  last  car. 

10.  See  page  158.  15.  123  volts  at  last  car. 

11.  444.38  volts  at  4th  car.  16.  487.6  volts  at  opposite  corner. 

12.  617.26  volts.  17.  5.4  volts  at  opposite  corner. 

18.  79.23  volts  at  car  on  opposite  corner. 

19.  454.47  volts  at  car  on  opposite  corner. 


CHAPTER   VII 

1.  10  kilowatts ;  13.45  horse  power.  4.  200  watts. 

2.  4.5  watts.  5.  300  watts. 

3.  55  watts;  3.44  watts  per  candle.    6.  15.41  horse  power. 
8.  86.96%.  9.  50%.  10.  80%. 

11.  Series         82.3  watts,  12.  Series         11.75  watts, 

12  groups  57.6      »  12  groups  23.04      « 

8      «       38.4      «  8      «       23.04      « 

6      «       26.2      «  6       "        19.00      " 

4      «       13.7      «  4      «        11.70      « 

3  «         8.2      «  3      «         7.50      « 

2  «         3.84    «  2      «          3.69      « 
multiple      .99  watt.  multiple       .978  watt. 

13.  Series         14.30%,  14.  Series         14.30%, 

12  groups  40.00%,  12  grpups  40.00%, 

8      «        60.00%,  8      ""        60.00%, 

6      «       72.72%,  6      «        72.72%, 

4  «       85.70%,  4      «        85.70%, 

3  «        91.43%,  3      «        91.43%, 
2      «        96.00%,                                     2      «        96.00%, 
multiple  98.97%.  multiple  98.97%. 

15.  See  page  158. 

16.  25%,  4346  watts ;  50%,  5390  watts  ;  75%,  7134  watts  ;  100%,  9578  watts ; 

125%,  12,722  watts. 


148 


ELECTRICAL  PROBLEMS 


17.  25%,  85.20%;  50%,  90.30%;  75%,  91.30%;  100%,  91.26%;  125%,  90.! 


18.  69.25  volts.  20.  125.75  volts. 

19.  84.49%.  21.  93.04%. 

24.  Dynamo  7.8% ;  line  10% ;  arcs  82.2%. 

25.  90.6%.  27.  24.2  amperes. 

26.  90.41  amperes.  28.  43.05  amperes. 


22.  86%. 

23.  140.75  volts. 

29.  24.5  horse  power. 


CHAPTER   VIII 

6.  .656  gauss. 

7.  0. 

8.  1.968  gausses. 

9.  1.26  gausses. 
10.  10  amperes. 

16.  .0716  ampere;  a  sine  galvanometer. 

17.  .143  ampere.  20.  18,312  gausses. 

18.  75.4  gausses.  21.  107.4  gausses. 

19.  122.145  gausses.         22.  82.4  gausses. 
25.  13,755  gausses ;  43,213  maxwells. 


1.  .4  gauss. 

2.  .48  gauss. 

3.  2.18  gausses. 

4.  .1  gauss. 

5.  1.136  gausses. 


26.  37.7  centimeters. 

27.  .02512  centimeter. 
32.  A,  20,000  gausses  ; 

B,  11,428 

C,  12,245        " 

D,  6,000 

E,  8,333        « 

37.  106.56  x  10 ~8  coulomb 
151.8    x  10-8 
116.5     x  10-*        " 


28.  .2528  centimeter. 

29.  8  maxwells. 

33.  A,  2540  gilberts  ; 

B,  136.3  gilberts; 

C,  397  gilberts; 

D,  1910.8  gilberts  ; 

E,  6250  gilberts. 

38. 
39. 


11.  1.5  centimeters. 

12.  .196  gauss. 

13.  48°  17'  .5. 

14.  .137  gauss. 

15.  .143  ampere. 

23.  9.17  gausses. 

24.  28.8  maxwells. 


30.  502.4  gilberts. 

31.  125.7  gausses. 

34.  2.38  amperes. 

35.  1.06  amperes. 

36.  18,  53,280 ; 

21,  75,  917; 

22,  58,245. 

6.747  x  10~8  coulomb. 
166.6  x  10~8  coulomb. 


40.  3710.5  divisions. 


CHAPTER   IX 

1.  Ll  =  .014  henry,  L2  =  .00014  henry. 

2.  Ll  =  7.56  henrys,  Z2  =  .0756  henry. 

3.  Zx  =  3.755  henrys,  Z2  =  .03755  henry. 

4.  .5445  henry,  .0667  henry. 

5.  Zj  =  .03447  henry,  Lz  =  13.79  henrys. 

6.  Zj  =  .158  henry,  Z2  =  .000253  henry. 


7. 


=  1.257  henrys,  L2  =  .0503  henry. 


ANSWERS  149 

8.  1.672  henrys.  10.  See  page  157. 

11.  .825  ampere;  82.43  amperes;  216.15  amperes;  250  amperes. 

12.  T  -  .04  ;  7.584  amperes.  13.  See  page  157. 

14.  See  page  157.  17.  .000063  henry.  20.    .0049  henry. 

15.  See  page  157.  18.  .000474  henry.  21.    .0625  henry. 

16.  17.2  henrys.  19.    .579  henry.  22.    .059  henry. 

23.  .251  henry.  24.  M  -  .00798  henry;  L  =  .1825  henry. 

25.  .0019  henry.  26.  .704  henry.  27.  .0035  henry. 


CHAPTER   X 

1.  100,000  C.G.S.  units.  2.  250.  3.  20  C.G.S.  units. 

4.  .00000006  coulomb.  5.  .0000000504  second. 

6.  7958  C.G.S.  units  or  .00884  microfarad.  7.  61  plates. 

8.  9804.6  square  centimeters.  9.  5.22  microfarads. 

10.  316,626  C.G.S.  units,  .3519  microfarad.  11.  39  plates. 

12.  .010063  microfarad.  14.  159;  17.37  centimeters. 

13.  7.95  centimeters.  15.   1.002  microfarads. 

16.  2.445  microfarads.      17.  .186  microfarad.  \8.  See  page  157. 

19.  See  page  157.  21.  13.817  x  RC  seconds. 

22.  .0004424  coulomb ;  .001554  coulomb;  .001987  coulomb. 

23.  .3894  ampere;  .1116  ampere;  .0034  ampere. 

24.  T  =  .00025  ;  q  =  .0000237  coulomb  ;  i  =  .0552  ampere. 

25.  See  page  157.  29.  .00543  microfarad.    33.  9  microfarads. 

26.  See  page  158.  30.  .00927  microfarad.     34.  .923  microfarad. 

27.  See  page  158.  31.  .521  microfarad.         35.  2.22  microfarads. 

28.  See  paga  159.  32.  1.199  microfarads. 


CHAPTER   XI 

2.  .000167  volt;  current  leaves  the  lead.  4.  .00156  volt, 

3.  .00178  volt;  current  towards  the  iron.  5.  See  page  159. 

6.  See  page  158.  9.  835°  C.  12.  977°  C.,  458°  C. 

7.  See  page  159.  10.  .00000477  ampere. 

8.  .00000116  ampere.     11.  27.3  centimeters. 


150  ELECTRICAL  PROBLEMS 


CHAPTER  XII 

2.  Hydrogen,  .0000416  gram  per   second;    chlorine,  .001468  gram  per 

second. 

3.  1.18  grams.  4.  33.87  amperes;  22.758  grams. 

5.  49  minutes,  1  second.  6.  10  hours,  47  minutes,  22  seconds. 

7.  17.83  amperes,  2.17  amperes  ;   oxygen  .00148  gram  per  second,  hydro- 

gen .000185  gram  per  second. 

8.  538.86  tons.  9.  204  days,  3  hours,  39  minutes,  32  seconds. 
10.  6.53.                       11.  1.2  amperes. 


CHAPTER   XIII 

1.  See  page  160.                5.  See  page  161.  9.  See  page  159. 

2.  7.07  volts.                      6.  See  page  159.  10.  See  page  161. 

3.  See  page  160.                7.  See  page  159.  11.  See  page  159. 

4.  3.536  amperes.              8.  See  page  159.  12.  See  page  160. 

13.  See  page  161.  14.  Virtual  value  =  71.25  volts;  see  page  163. 

15.  Virtual  value  =  72.4  amperes ;  see  page  161. 


CHAPTER   XIV 

1.  132.9  volts.  2.  135.3  volts.  3.  121.7  volts. 

4.  30°,  193.2  volts;   60°,  173.2  volts;  90°,  141.4  volts;  120°,  100  volts; 

150°,  51.8  volts  ;  180°,  0. 

5.  90°.  6.  36°  52'.          7.  62°  43',  26°  23'.  8.  103.1  volts. 
9.  141.8  volts,  81°  56'.           11.  85.4  volts,  114°  11'.          12.  0. 

13.  1732  volts  ;  phase  angles  90°,  210°,  330° ;  see  Problem  13,  Chapter  XIII. 

14.  1000,  0° ;  1414,  45°  ;  1000,  90°.  15.  707.5  volts. 

CHAPTER   XV 

1.  1.414  : 1.000.  2.  50  amperes.  3.  0. 

4.  13  amperes;  its  negative  is  22°  37'  away  from  12  ampere  component 
in  phase.  5.  2.65  amperes. 

7.  Leading  current,  18.42  amperes;  lagging  current,  7.57  amperes. 

8.  3.47  amperes.  11.  158.81  amperes,  74°  50'. 

9.  14.38  amperes,  76°,  51'.  12.  173.2  amperes. 
10.  104.54  amperes,  247°  30'.  13.  .174  ampere. 


ANSWERS  151 


CHAPTER   XVI 

2.  .003  henry.  3.  36°  52'.  4.  See  page  162.  5.  5.9  ohms. 

6.  11.7  amperes.  7.  200  volts.  8.  182.3. 

9.  Problem  7,  22°  31'  ;  Problem  8,  66°  25'.  10.  .102  henry. 

11.  18.39  ohms.  22.  254.6  microfarads.  33.  3432  ohms. 

12.  266.4  ohms.  23.  7.69  amperes.  34.  363.2  ohms. 

13.  9.165  ohms.  24.  22°  37'.  35.  133  ohms. 

14.  8.62  ohms.  25.  See  page  164.  36.  See  page  163. 

15.  44.77  ohms.  26.  376  ohms.  37.  See  page  163. 

16.  27.4  ohms.  27.  2.18  microfarads.  38.  See  page  163. 

17.  62.25.  28.  5804  ohms.  39.  15  ohms. 

18.  See  page  163.  29.  Increased  79.6%.  40.  6.67  amperes. 

19.  See  page  163.  30.  Decreased  33.4%.  41.  36°  52'  lag. 

20.  See  page  162.  31.  Decreased  10.2%.  42.  See  page  164. 

21.  13  ohms.  32.  Increased  33.2%.  43.  23.89  ohms. 

44.  209  ohms.  48.  Current  leads  48°  54'. 

45.  Current  leads  30°  30'.  49.  See  page  165. 

46.  5.21  henrys.  50.  See  page  165. 

47.  157.9  amperes.  51.  See  page  165. 

52.  See  page  166.  53.  Condenser  990  volts,  inductance  1000  volts. 

54.  See  page  165.  56.  See  page  165. 

55.  See  page  165.  57.  See  page  165. 

58.  E.M.F.  =  106.255  volts ;  61.  7X  =  29.54  amperes ; 
resistance  =  4.082  ohms ;  72  =  15.86  amperes ; 
reactance  =  1.182  ohms;                       73  =  22.83  amperes ; 
impedance  =  4.25  ohms  ;  74  =  7.8  amperes ; 

II  —  21.65  amperes,  lags  35°  15'  resistance  =  175.2  ohms ; 

behind  the  E.M.F. ;  reactance  =  102.97  ohms  ; 

72  =  8.105  amperes,  leads  E.M.F.  impedance  =  203.2  ohms ; 

by  23°  50' ;  7  =  54.13   amperes,  lags  30°  25' 

7  =  25  amperes,  lags  16°  9'  behind  behind  the  E.M.F. 

the  E.M.F.                                   62.  7  =  .19864  ampere  in  phase  with 

59.  Resistance  =  353.55  ohms  ;  the  E.M.F. ; 

reactance  =  0  ;  impressed  E.M.F.  =  100.5  volts ; 

impedance  =  353.55  ohms ;  resistance  =  505.9  ohms ; 

current  =  2.828  amperes  in  phase  reactance  =  0 ; 

with  the  E.M.F.  impedance  =  505.9  ohms. 

60.  See  page  167. 


152  ELECTRICAL  PROBLEMS 


CHAPTER   XVII 

2.  40  volts.       3.  53  volts.       4.  187.5  R.P.M.      5.  11.8%.       6.  314  volts. 

7.  566  volts.     8.  25.73  megamaxwells.  9.  6.96  kilogausses. 

10.  8.21  kilogausses.  11.  .87  square  centimeter. 

12.  96.  15.  609.33  volts.  18.  1.76  megamaxwells. 

13.  6  poles,  3.55%.  16.  6186  gausses.  19.  12  kilogausses. 

14.  30.  17.  No.  6.  20.  324.65  volts. 
21.  282.2  amperes  per  square  centimeter.  22.   .01  ohm. 

23.  291  centimeters.  28.  3.253  megamaxwells. 

24.  12,305  gausses.  29.   1.705  megamaxwells. 

25.  764  square  centimeters,  30.  533  square  centimeters. 

26.  275  R.P.M.  31.  10.1  megamaxwells. 

27.  16.1  volts. 


CHAPTER   XVIII 

1.  746  volts.  3.  457.1  volts.  5.  7,109  gausses. 

2.  2390  volts.  4.  2. 98  megamaxwells.     6.  7281  volts. 

7.  98.93  R.P.M. 

8.  1200  R.P.M.;  .07  megamaxwell. 

9.  3554  volts.         10.  125%.        11.  27.3%.         12.   1.619  megamaxwells. 
13.  5.44%.  14.  No.  10.          15.   1.72  megamaxwells. 

16.  80  R.P.M.  17.  577.3  kilowatts.      18.  318.      19.  225 


CHAPTER   XIX 

2.        ONE  WINDING  TABLE  4.          ONE  WINDING  TABLE 

Com.           B          F           Com.  Com.           B          F 

1  1           12              2  118 

2  3          14             3  2              3          10             3 

3  5          16             4  3              5          12             4 

Etc.  Etc. 

5.        ONE  WINDING  TABLE  7.      Com.          B         F           Com. 

Com.          B          F          Com.  118               8 

1  1          24             2  8            15        22             15 

2  3          26             3  15            29        36             22 

3  5          28             4  Etc. 

Etc. 


ANSWERS 


153 


8.        ONE  WINDING  TABLE              9. 

Com. 

Com.          B         F 

Com. 

1 

I              1        14 

14 

16 

14            27        40 

27 

31 

27            53        12 

18 

Etc. 

10.  160  or  164. 

11. 

HAND  AVINDING 

First  Layer 

1          13 

25 

Second  Layer     14          26 

38 

12. 

MACHINE  WOUND  COILS 

1 

22               43 

64 

3 

24               45 

66 

Etc. 

HAND  WINDING 

First  Layer            1 

21               41 

61 

Second  Layer    105 

2               22 

42 

Third  Layer       86 

106                 3 

23 

Fourth  Layer      67 

87             107 

4 

F 

16 


31        46 

61        76 

Etc. 


37    Etc. 
4    Etc. 


85 
87 


106 
108 


Com. 

16 

31 

2 


81  101  Etc. 

62  82  Etc. 

43  63  Etc. 

24  44  Etc. 


14.  Twelve  poles,  144  conductors  in  72  coils,  multiple  grouping. 

15.  Six  poles,  56  conductors  in  28  coils,  series  grouping. 

16.  Six  poles,  120  conductors  in  60  coils,  multiple  grouping. 

17.  Four  poles,  78  conductors  in  39  coils,  series  grouping. 

18.  Fourteen  poles,  616  conductors  in  154  coils,  multiple  grouping. 


CHAPTER  XX 


5.  272  ampere  turns. 


1.  1200  per  pole.  3.  480  per  pole. 

2.  489  per  pole.  4.  3°.23. 

6.  Drop  due  to  resistance  is  39.27  volts;  drop  due  to  inductance  is  247.9 

volts;  inductance,  .00683  henry. 

7.  81°.  8.  2053.  9.  1040.  10.  408. 


1.  2427  ampere  turns. 

2.  110  ampere  turns. 

3.  643  ampere  turns. 


CHAPTER   XXI 

4.  136  ampere  turns. 

5.  419  ampere  turns. 

6.  3735;  14,940. 


7.  No.  21. 

8.  4.66  centimeters. 


154  ELECTRICAL  PROBLEMS 

9.  977.5  ohms  at  40°  C. ;  1045  ohms  at  60°  C. 

10.  80  ohms  at  60°  C. ;  147.5  ohms  at  40°  C. 

11.  278  watts  at  40°  C. ;  297  watts  at  60°  C. 

12.  19°.23  C.  14.  591  ampere  turns.         .  16.  No.  20. 

13.  15,992  ampere  turns.         15.  4589  ampere  turns.         17.  29°.5  C. 

18.  87  turns  per  pole  of  conductor  with  an  area  of  .0833  square  centimeter. 

19.  6764  ampere  turns.          20.  135.28  amperes.  21.  .52  ohm. 

22.  .0021  henry  per  phase.  24.  154.42  amperes. 

23.  35.4  amperes.  25.  178.92  amperes. 


CHAPTER   XXII 

2.  60.       3.  11,000  volts;  2000  convolutions.      4.  1950  volts.       5.  15.7%. 

6.  Current  =  4.2  amperes ;  E.M.F.  =  2042.6  volts,  current  lags  53°  13'. 

7.  Current  =  4.011  amperes  ;  E.M.F.  =  1990.3  volts,  current  leads  33°  25'. 

8.  Current  =  5.727  amperes  ;  E.M.F.  =  2023.5  volts,  current  lags  3°  27'. 

9.  .16  ampere,  206  watts.  10.  .62. 

11.  Primary,  2.19  ohms;  secondary,  .00464  ohm. 

12.  Primary,  106.9    amperes   per   square    centimeter  ;    secondary,   109.24 

amperes  per  square  centimeter. 

13.  Primary,  16  volts;  secondary,  .67  volt.  14.  214  watts. 

15.  I  load,  94.46%;  \  load,  96.65%;  f  load,  97.18%  ;  full  load,  97.28%. 

16.  \  load,  94.3%;  \  load,  96.4%;  f  load,  96.8%  ;  full  load,  96.8%. 

17.  \  load,  1105.7  volts;  \  load,  1110.6  volts;  f  load,  1115.4  volts;  full 

load,  1120.3  volts.  18.  10,316  gausses. 

19.  12.604  kilowatts.  20.  18.06  kilowatts. 

21.  £  load,  97.56%  ;  £  load,  98.47%;  f  load,  98.64%  ;  full  load,  98.63%. 

22.  See  page  163.  23.  98.68%.  24.  96.56%.  25.  95.16%. 
26.  About  3500  volts.     The  secondary,  taking  full  load  current,  becomes 

the   primary.     This   current   sets  up  a  very  large    magnetic   flux 
through  the  primary,  which  produces  a  large  electro-motive  force. 


CHAPTER   XXIII 

1.  35.35  amperes  in  each  circuit.  2.  336.8  volts.            3.  94.3  amperes. 

4.  84.7%.                   5.  83.8%.  6.  40.06  amperes.      7.  10.9  amperes. 

8.  900  revolutions.  11.  91.5%  at  200  kilowatts  ; 

9.  2688  armature  conductors.  95.2%  at  400  kilowatts; 
10.  5317  watts.  96.3%  at  600  kilowatts. 


ANSWERS 


155 


CHAPTER   XXIV 


1.  1200R.P.M. 

2.  217,280  maxwells. 

3.  4.2  amperes. 

4.  .45  ampere. 

5.  .006  ohm. 

6.  99.2  amperes. 
18. 

19. 


7.  104.3  amperes. 

8.  4.75%. 

9.  .000036  henry. 

10.  000655  ohm. 

11.  6°  ir. 

12.  7.5 ;   13.9  amperes. 

Cos  0  =  .88  ;  real  power  -=-  apparent  power  =  .88  +  . 
81.7%.        20.  4.62  amperes;  power  factor  =  .25.       21. 


13.  15.38  amperes. 

14.  .427  ohm. 

15.  303  watts. 

16.  .832  ohm. 

17.  110.26  volts. 


23.  .89. 

24.  0  load  =    8.0  amperes  ; 
5  load  =  10.4  amperes ; 
|  load  =  14.9  amperes ; 
J  load  —  20.0  amperes ; 

full  load  =  26.6  amperes. 

25.  .1177  ohm. 


26.  0  load,     5.8  amperes; 
£  load,    7.2  amperes  ; 
I  load,    9.3  amperes; 
|  load,  12.8  amperes ; 

full  load,  16.7  amperes. 

27.  See  page  168. 

28.  See  page  169. 


22.  .83. 


CHAPTER   XXV 

1.  70%.  2.  4.31%.  3.  32%. 

4.  A  =  .15;   C=  6  x  10~8; 

field  current  2.5  amperes,  a  =  .07,  B  =  .00004  ; 

field  current  5.0  amperes,  a  =  .18725,  B  =  .0001336  ; 

field  current  7.5  amperes,  a  =  .2825,  B  =  .000229. 

5.  1041  watts.  6.  82.4%.  7.  80%. 

8.  T  -  1.24,  flux  each  leg  =  3,436,700  maxwells;  leakage  coefficient  =  1.83. 

9.  See  page  170.  10.  See  page  170. 


1.  See  page  167. 

2.  See  page  165. 

3.  See  page  167. 

4.  56.1  miles. 

5.  $22.14. 


CHAPTER   XXVI 

6.  10,895  volts. 

7.  22,230  volts. 

8.  48,744  volts. 

9.  72,790  volts. 
10.  72,837  volts. 


11.  2098  kilowatts. 

12.  10,000  volts. 

13.  See  page  167. 

14.  .9. 


15.  Ratio  of  weights  of  copper,  single  phase  1.000,  2  phase  1.457,  3  phase 

.750. 

16.  .268  inch.  17.  Efficiency,  87.8% ;  power  factor,  .97. 


GEAPHICAL   SOLUTIONS 


3-50 

£.25- 


0°  100°  200° 

Temperature, 


300° 


Problem  27,  Chapter  IV 


2  4- 

^^ 

"^  2 

f 

1'- 
<3o 

',,,,, 

2             4             G              8             1( 

Time  Seconds 

Problem  10,  Chapter  IX 


I7'5 
1  5- 
1 2.5 

0- 


4  8  12  16 

Inductance  Henrys 

Problem  13,  Chapter  IX 


7?  =  100 


Time  Seconds 
Problem  15,  Chapter  IX 


•jo 


2  4  68 

Time  Seconds. 

Problem  14,  Chapter  IX 


.001         .062 

Ttm-e 


.003 


Problem-19,  Chapter  X 


.004 


100        200        300        400        500 

Resistance 


157 


Problem  25,  Chapter  X 


158 


ELECTRICAL   PROBLEMS 


no 


I  105 


Problem  8 


100 


500  600  700  800  900  1000  1100 

Distance  in  feet  from  Generator 

Problem  10,  Chapter  VI 


1200     1300     1400 


lOOr   lOOr 


10          20          30          40          50          60          70          80          90         100 
Per  Cent  of  Total  Resistance  in  External  Circuit 
Problem  15,  Chapter  VII 


<  =  .0001   Second 

<-.0005  Second 

<-.002    Second 


.o  .0005 
I  .0004 
6  .0003 
|.0002 
g.OOOlf 


-  r=.0001  Second 
---  <=.0005  Second 
-----  1=.002  Second 


12345 
Capacity  Microfarads 


Problem  27,  Chapter  X 


100         200         300         400        500  Cq 

Resistance  Ohms 


'0°     100°  200°    300°    400°   500°    600° 
Temperature 


Problem  26,  Chapter  X 


Problem  6,  Chapter 


GRAPHICAL    SOLUTIONS 


159 


100 

50- 

0 

60 

100 


150 
100 
50 


bj  0 

^60 

.100 

150 


1234 
Capacity  Microfarads 

Problem  28,  Chapter  X 


.OJL25Scc. 


Problem  6,  Chapter  XIII 


Problem  7,  Chapter  XIII 


.00625  .0125   .01875  .0250  Sec. 


Problem  5,  Chapter  XI 


.015- 


0  500°  1000°  1500° 

Temperature 

Problem  7,  Chapter  XI 


25  Sec.  75 

2  50 
1 25 

?    °' 
f  25 

°  50 
75 


,100 
\   50 


f  .0025        .0050      .0075  ^OJOO  Sec 


Problem  9,  Chapter  XIII 


:  50 
;ioo 


.0050        .U100      .0150    . 


Problem  8,  Chapter  XIII 


Problem  11,  Chapter  XIII 


I  GO 


ELECTEICAL   PROBLEMS 


Problem  1,  Chapter  XIII 


Problem  3,  Chapter  XIII 


01666  -Sec- 


Problem  12,  Chapter  XIII 


GRAPHICAL    SOLUTIONS 


161 


Problem  5,  Chapter  XIII 


2000 

Problem  13,  Chapter  XIII 


04  Sec. 


£  50 
"o 
^   25 

N   ° 

fcq    25 
50 

75 


Problem  10,  Chapter  XIII 


40  80  120  1GO  200  240  280  320  3GO 


Problem  15,  Chapter  XIII 


162 


ELECTKICAL  PKOBLEMS 


75 


50 


25 


50- 


75 


Problem  4,  Chapter  XVI 


10 


150 


.2  .3 

L  in  Henrys   R  =  50 


200 


Problem  20,  Chapter  XVI 


GRAPHICAL   SOLUTIONS 


163 


^100 
1  75 
kj  50 
^  25 
0 


0°        30°      60°      90°      120°    150 

Problem  14,  Chapter  XIII 


.2  .3 

Inductance  Henry 

Problem  18,  Chapter  XVI 


300 


200 


100 


50  100  150 

Resistance  Ohms 


Problem  19,  Chapter  XVI 


200 


1200 


400 


300          600  900 

Resistance  Ohms 

Problem  36,  Chapter  XVI 


1200 


1200 


§1000 


600 


0246 

Capacity  Microfarads 

Problem  37,  Chapter  XVI 


20 


1,10 


H=  100  Constant 


25  50  75  100 

Capacity  Microfarads    72  =  100  Const. 


250  500  750  1000 

Resistance  Ohms        C=10  Const. 

Problem  38,  Chapter  XVI 


10 


20 


-10 


-5, 


jo 


10 


e- 


io 


'20  -'IS  5 

Problem  22,  Chapter  XXII 


164 


ELECTRICAL  PKOBLEMS 


10 


150 


Problem  25,  Chapter  XVI 


1,7 


1(1 


150 


tire 


L5( 


Problem  42,  Chapter  XVI 


GRAPHICAL    SOLUTIONS 


165 


2  2000 

s 

<r 

c 
s  1000 


05  10  15 

Capacity  Microfarads 

Problem  49,  Chapter  XVI 


1600 


-51200 


"S  800 

a, 


400 


01234 
Inductance  Henrys 

Problem  50,  Chapter  XVI 


S7600- 


7500 


250  500  750 

Resistance  Ohms 

Problem  51,  Chapter  XVI 


1000 


500 


1000 


10    20   30  40   50   GO    70  80   90   100 
Capacity  Microfarads 

Problem  54,  Chapter  XVI 


1000 


.1    .2    .3    .4    .5    .0    .7    .8   .9     I. 
^  Inductance  Henry 

Problem  55,  Chapter  XVI 


400 


300 


t  200 


100 


200        400        GOO        800       1000 
Resistance  Ohms 

Problem  56,  Chapter  XVI 


1000 


Z-coJ 


0  100  200  300  400 

Frequency  Cycles  Per  Second  . 

Problem  57,  Chapter  XVI 


20          40          GO          80          100 
Length  of  Line  Miles 

Problem  2,  Chapter  XXVI 


166 


ELECTRICAL   PROBLEMS 


10 


20 


30  40  50  60          70 

Capacity  Microfarads  Curve  I 


SO 


90 


3456- 
.Inductance  Henrys  Curve,  11 


10 


100         200 


300         400         500         600         700 
Resistance  Ohms  Curve.lII. 


800          900 


1000 


10  15  20  25  30  35 

frequency  Cycles  Per  Second  Curve  IV 


50 


Problem  52,  Chapter  XVI 


GRAPHICAL   SOLUTIONS 


1G7 


4000- 


3000- 


2000- 


1000- 


2  4  68  10 

Capacity 
Problem  CO,  Chapter  XVI 


12 


16 


4000 


804 


r 


11,000- 


10,000- 


9,000 


2000 


1000 


25$  50 f°  757 

Efficiency  Per  Cent 

Problem  1,  Chapter  XXVI 


2000       4000        6000        8000       10000 
Volts     at     Generator 

Problem  3,  Chapter  XXVI 


125 


250 


375  500 

Load  in  Kilowatts 


625 


750 


Problem  13,  Chapter  XXVI 


1G8 


ELECTRICAL  PROBLEMS 


Amperes    Input 


Torque  in  3feter  Kilograms 


Problem  27,  Chapter  XXIV 


GRAPHICAL    SOLUTIONS 


1.09 


Ampere*  Input 


Torque  in  Meter  Kilograms 


Slip  Per  Cent 
Problem  28,  Chapter  XXIV 


170 


ELECTRICAL  PROBLEMS 


60 


40- 


20 


)                 100 

200 

300               400               500 

Speed  Revolutions  per  Minute    , 

600 

700 

800 

25 

50        Speed  Mii^per  Hour        10° 

125 

150 

500 

1000 

1500              2000             2500 
Torque  Lbs.  on    One  Foot  Radius 

3000 

3500 

4000 

400 

800 

1200              1600              2000 
Tractive  Effort  Pounds 

2400 

2800 

3200 

20 

40 

60                  80                 100 

120 

40 

160 

Efficiency  Per  Cent 


Problem  9,  Chapter  XXV 


100 


80- 


GO 


2 

4 

6 

8                 10 
Miles  Per  Hour 

12 

14 

16 

18 

0 

250 

500 

750 

T^veE.tff 

1500 

1750 

2000 

2250 

0 

10 

20 

30 

40                50 
Efficiency  per  Cent 

CO 

70 

80 

90 

Problem  1,0,  Chapter  XXV 

- 


OF  THE          '>A 

UNIVERSITY  j 


UNIVERSITY  OP  CALIFORNIA  LIBRARY 


THIS  BOOK  IS  DUE  ON  THE  LAST  DATE 
STAMPED  BELOW 


OCT  7  1916 


nr 


JAN    8 


0£C  20  1923 
DEC    2  1927 


- 


'  3 
9 


MOV  8  1946 


30m-l,'15 


#7 ft 


1 0367:5 


i 


